Q:

Winnie-the-Pooh and the 27 honey pots

Winnie-the-Pooh keeps his $27$ honey pots in the larder. 
Each pot contains up to $1$ kilogram of honey, and different pots 
contain different quantities of honey. 
All $27$ pots together contain $17$ kilogram of honey.
Every day, Winnie-the-Pooh selects $7$ pots, picks a real number $x$, 
and then eats exactly $x$ kilogram of honey from each of the selected pots.

Question: Is it always (independently of the initial distribution 
  of honey) possible for Winnie to empty all $27$ honey pots in a finite 
  number of days?

A:

Update: I made a big blunder in the previous version, this should be fixed now.

Part 1: It is always possible to remove amount $x$ from any number $n \ge 7$ of pots if all pots have at least amount $x$
Let $p$ be the number of permutations with 7 pots containing a specific pot. Cycle all permutations and remove amount $\frac xp$ from each pot. As each pot occurs $p$ times the total removed amount from each pot is $x$.
Part 2: A combination of pots where the 7 pots containing the most amounts are equal is know to work regardless of the amounts in other pots
Consider the following image:

The bars show the amounts in the pots, ordered by the amount. The amounts in the first 7 pots are equal. Start with the smallest pot $J$ and remove the amount of $J$ from all pots (this works as seen in part 1). Repeat with pot $I$ and then with pot $H$. At the end we have the top 7 pots containing only the black marked area, which can be emptied in the next step.
Part 3: Removing superfluous amount in biggest pot requires 6 times the same amount 20 smallest pots
Consider the following image:

We must remove the red part in pot $A$ to reach the situation from part 2. It is easy to see that we can do this easily if we have the same 6 parts in pots $H-M$.

This is the place where I made a blunder in the previous version, assuming I could trade the red part in $A$ with a combination of parts of the same size in pots $>G$, instead of 6 times the size.

What can be said about the distribution of the amounts beyond pot $G$? Actually it doesn't matter, if we use a trick. Consider the following image.

We want to remove the red part in pot $A$. The orange part in pot $H$ has 6 times the size of the red part. We can remove the red part and the orange part of pot $H$ while also removing the orange parts in pots $A-G$. To do this we generate all permutations of 6 pots from the range $B-H$:
$\binom{7}{6}$ = 7
The number each of the pots $B-H$ appears in this permutations is:
$\binom{6}{5}$ = 6
Now we add pot $A$ to all these permutations and remove the red part from each permutation. We remove the red part 7 times from pot $A$ and 6 times from pots $B-H$ which is exactly what we see in the image above. This can be easily extended to any number of pots, as long as they contain 6 times the size of the red part. 
The other conclusion is, no matter what we do, we can't remove the red part if the size of the pots beyond $G$ is less than 6 times the size of the red part.
Part 4: Removing superfluous amount in 2 biggest pot requires 2.5 times the same amount 20 smallest pots
Consider the following image:

Similar to part 3, we now want to remove the 2 red parts in pots $A-B$. It is again easy to see that we can do this easily if we have the same 5 parts in pots $H-L$.
Using the same reasoning as in part 3, it can be shown, that the amount distribution beyond pot $G$ doesn't matter.
It can also be shown that similar applies to superfluous amounts in pots $C-F$.
Part 5: The general case
The following image shows the general case which is guaranteed to be solvable based on the parts above:

As we have seen in part 3, the distribution in pots beyond $G$ doesn't matter. Also as seen in part 2 adding amounts in pots beyond $G$ is solvable too. What is left is to show which amounts can be solved in any case.
Part 6: The edge case

The above image shows the edge case, where the red area is maximized (if we limit ourselves to 13 pots). The height of pots $H-M$ must be the same as the height of the red part of pot $A$. Pots $B-G$ must have also the same size. As $A$ is limited to $1$, the height of pots $B-M$ must be $\frac{1}{2}$. The size of this case is:
$1 + 12 * \frac{1}{2} = 7$
This means that 7kg honey is enough to guarantee, that the puzzle is solvable, if the pot size is limited to 1kg.
Any other amounts which we would try to add, don't change anything. Amounts added to pots beyond $G$ are OK as shown in part 2. Amounts added to pots $B-G$ reduce the required amount in pots beyond $G$ as can be seen in the following image:

One more thing we could do, is to use the other 14 pots we didn't use yet to further maximize the red area, like in the following image:

We used here one more pot. We increased the size of the red area above $A$ by some amount $x$. We also increased the size of the red area beyond pot $G$ by $6x$. But at the sime time we decreased the size of the black area by $7x$ which in the end gives a total change of $0$.

The pot size which would still work for 17kg is (based on the first edge case):
$x + 12 * \frac{x}{2} = 17$
$x = \frac{17}{7} = 4,28...$
Also, this solutions works with any number of pots $\ge 13$.

Alternatively we can make Winnie-the-Pooh really happy and tell him, he can eat from 17 pots each day (which would still work with 17kg and 1kg max per pot).

A:

Here is a rigorous proof that Pooh can always succeed.
Let $P_i$ be the amount of honey in the $i^{th}$ pot when sorted in decreasing honey contents, so $1\ge P_1\ge P_2\ge \dots\ge P_{27}\ge0$ and $P_1+\dots+P_{27}=17$. 
Using the beginning of Sleafar's solution, we now need only show how to equalize the seven heaviest pots. At any point, let $m$ be the number of pots which have the same amount of honey as the heaviest, and let $n$ be the number of nonempty pots (so initially, $m=1$ and $n=27$). We show that, as long as $m<7$ and $n\ge 14$, it is possible to either increase $m$ or decrease $n$. We then argue why $n$ can't drop below $14$ before $m$ reaches $7$, so that eventually we equalize the top $7$ pots.
Here is the method. Let $x$ be the gap in honey contents between the heaviest $m$ pots and the next strictly lighter pot. Each day for the next $n-7$ days, Pooh eats $\frac{x}{n-7}$ kg from each of the $m$ heaviest pots. Each of these days, he must eat the same amount from some $7-m$ other pots. He does this in such a way that the $n-7$ pots $P_8,P_9,\dots,P_n$ are eaten from equally over this $n-7$ day period (for example, by arranging the nonempty pots in a circle, and eating from each consecutive segment of length $7-m$). 
As long as each of the nonempty pots contain at least $\frac{x}{n-7}\cdot (7-m)$ kg of honey, then this will work, causing the $m$ heaviest pots to decrease to the next highest weight, thus increasing the number $m$. If the nonempty pots do not contain enough honey to do this, then adjust $x$ so that the smallest pot contains exactly $\frac{x}{n-7}\cdot (7-m)$. Then doing the procedure will empty that pot, thus decreasing $n$.
Now, why can't $n$ drop below $14$ before $m$ reaches $7$? First off, the amount of honey that needs to be removed from pots 8 through $n$ in order to equalize pots 1 through 7 is equal to 
$$
6(P_1-P_2)+5(P_2-P_3)+4(P_3-P_4)+3(P_4-P_5)+2(P_5-P_6)+(P_6-P_7)
$$
This is because, during the phase when pots $P_1,\dots,P_m$ are equal, we have to remove a total of $P_{m}-P_{m+1}$ from each of the first $m$ pots, and $(7-m)$ times that from the last pots 8 through $n$. A different way of writing this is 
$$
6P_1-P_2-P_3-\dots -P_7\le 6 - 6P_7< 3
$$
The last inequality follows since $P_7> \frac12$. Why is this true? If not, then the first 6 pots would initially contain at most $6$ kg, and the last 21 would contain at most $\frac12$, so that the total amount of honey was at most $6+\frac12\cdot 21=16.5$, which is impossible since we started with $17$ kg of honey. 
So, the procedure will remove less than $3$ kg from the last 20 pots. However, the first 14 pots contain at most 14 kg, meaning the lightest pots together contain at least $17-14=3$ kg. Since our procedure eliminates the lightest pots first, the first 14 pots will not be emptied, as claimed.

A:

YES, HE CAN!

Rigorous proof and exact bounds:
  If the starting amount of honey in each pot is less or equal to $17/7$ kilograms, then Pooh can always do this. If the starting amount of honey in each pot can be more than $17/7$ kilograms, then Pooh can not always do this.
More generally: If the total amount of honey is $A$, the number of pots is $X$, each of the pots contains at most $P$ kilograms of honey and Pooh chooses every day $r \leq X$ pots, then he can always empty them if and only if $P \leq A/r$. (I have provided computations for $A=17, X=27, r=7$, but the same approach works for all triplets $(A, X, r)$.

Proof:
PART 1.

The idea is to arrange the pots by amount of honey in decreasing order - #1, #2, #3, etc. and then take honey in such way so that we get more and more pots with the same amount of honey as pot #7.

Arrange the pots by amount of honey:
$x_1\geq x_2\geq x_3\geq x_4\geq...\geq x_{27}$.
Assume that we have:
$x_1\geq x_2\geq ...\geq x_k>x_{k+1}=x_{k+2}=...=x_l>x_{l+1}\geq x_{l+2}\geq ...\geq x_{27}$,
where $k<7$ and $l\geq 7$ (this means that all pots with numbers from $k$ up to $l$ have the same amount of honey as pot #7). Now Pooh takes honey from the first $l$ pots, such that the amount of honey taken from each of pots $k+1, ... , l$ is $(7-k)/(l-k)$ of the amount of honey taken from each of pots $1,2,...,k$ and one of these two conditions is satisfied:

the amount of honey in pot $k$ becomes equal to the amount of honey in pots $k+1,k+2,...,l$;
the amount of honey in pot $l+1$ becomes equal to the amount of honey in pots $k+1, k+2,...,l$.

It is easy to see that Pooh can complete this task in $l$ days - just take honey from pots $1$, $2$, ... ,$k$ + some honey from $7-k$ pots among $k+1$,$k+2$,...$l$ in cyclic manner. After he completes this step, he repeats the process until all pots get equal amount of honey in them and eventually become empty.
Now we prove that eventually we will get $x_1=x_2=...=x_{27}=0$. Indeed, if we assume that Pooh gets stuck, then all of the pots except for at most $6$ should become empty. Let the number of non-empty pots be $m \leq 6$ and the total amount of honey remaining is $H \in [0,\min(mP,17))$. Clearly, the amount of honey taken from these $m$ pots is exactly $m(17-H)/7$ (because every time we take out honey, $m/7$ of it is taken out of them). Since each pot contains in the beginning at most $P$ kilograms of honey, we have
$$\frac{(17-H)m}{7}+H\leq mP.$$
The inequality above just states that the total amount of honey remaining in the last $m$ pots + the amount of honey taken out of them is no more than the maximum amount of honey there was in the pots initially. 
We can rewrite the inequality as:
$$0<H(7-m)\leq m(7P-17) \rightarrow P>17/7$$.
Therefore we can conclude that if $P\leq 17/7$, then Pooh can always do this.
PART 2

The idea is to fill up the first pot entirely.

Now we prove the inverse. Let the first pot has $P$ kilograms of honey and the others have $(17-P)/26$ each. Now every time we take $x$ kilograms of honey from the first pot, we take at least $6x$ kilograms of honey from the remaining pots. In order to empty the first pot, we have to take out $P$ kilograms of honey from it and therefore we have to take out $6P$ kilograms of honey from the other pots. However, $6P>17-P$, which is a contradiction.
Remark: If the total amount of honey is $A$ and the number of pots Pooh can use on every day is $r$, then using the same calculations it is easy to see that the optimal $P=A/r$. The total number of pots does not matter at all, it could be anywhere from $7$ to infinity.

