Combinatorics problem on $k$ sets and partionning a set of numbers consider $2k+1$ real numbers satisfying the property such that any $2k$ of them can be partitioned into two $k$ sets which exist that the sum of the numbers in each of the sets are equal. 
Prove that all the numbers are equal.
First you can see easily all numbers are even or all numbers are odd. If you take $a+k_1+k_2$ is even, its mean a is even because $k_1=k_2$ thus $k_1 + k_2 = 2\cdot k_1$ which is even. The sum $a+ k_1 +k_2$ is the sum of all numbers and must be all time even or all time odd. Let say $a$ was odd, than the hole sum is odd, so when you take $b + k'_1+k'_2$ you see $b$ is also odd.

Now regardless to that, lets say they are not all equal. So there is at least one pair on numbers such that $a &lt; b$. Lets take the ones that minimize the subtraction, I mean look for the minimum $a-b$ while $a-b&gt;0$. Now you get $a, k_1=k_2$ and $b,k'_1=k'_2$. 

Now I may not explain it well, and English is not my native so excuse me..
In the first configuration, b is in one of the groups $k_1$ or $k_2$, lets say $b\in k_1$. Swap $b$ and $a$ so you get the configuration $b, k_1\cup a,k_2$. Now try to move numbers from $k_1 \cup a$ to $k_2$ and in the other way to make them into 2 groups equal to each other (suppose to get $k'_1$ and $k'_2$ or other configuration that can hold it).
But because you took $a-b$ to be the minimum, you won't be able to do it. The proof from here is on you :)
First you can see easily all numbers are even or all numbers are odd. If you take $a+k_1+k_2$ is even, its mean a is even because $k_1=k_2$ thus $k_1 + k_2 = 2\cdot k_1$ which is even. The sum $a+ k_1 +k_2$ is the sum of all numbers and must be all time even or all time odd. Let say $a$ was odd, than the hole sum is odd, so when you take $b + k'_1+k'_2$ you see $b$ is also odd.

Now regardless to that, lets say they are not all equal. So there is at least one pair on numbers such that $a &lt; b$. Lets take the ones that minimize the subtraction, I mean look for the minimum $a-b$ while $a-b&gt;0$. Now you get $a, k_1=k_2$ and $b,k'_1=k'_2$. 

Now I may not explain it well, and English is not my native so excuse me..
In the first configuration, b is in one of the groups $k_1$ or $k_2$, lets say $b\in k_1$. Swap $b$ and $a$ so you get the configuration $b, k_1\cup a,k_2$. Now try to move numbers from $k_1 \cup a$ to $k_2$ and in the other way to make them into 2 groups equal to each other (suppose to get $k'_1$ and $k'_2$ or other configuration that can hold it).
But because you took $a-b$ to be the minimum, you won't be able to do it. The proof from here is on you :)