Number of possible sets for given N How many possible valid collections are there for a given positive integer N given the following conditions:
All the sums from 1 to N should be possible to be made by selecting some of the integers. Also this has to be done in way such that if any integer from 1 to N can be made in more than one way by combining other selected integers then that set of integers is not valid.
For example, with N = 7,
The valid collections are:{1,1,1,1,1,1,1},{1,1,1,4},{1,2,2,2},{1,2,4}
Invalid collections are:
{1,1,1,2,2} because the sum adds up to 7 but 2 can be made by {1,1} and {2}, 3 can be made by {1,1,1} and {1,2}, 4 can be made by {1,1,2} and {2,2} and similarly 5, 6 and 7 can also be made in multiple ways using the same set.
{1,1,3,6} because all from 1 to 7 can be uniquely made but the sum is not 7 (its 11).

Suppose your collection contains a finite set of distinct numbers $\{n_1 ... n_k\}$ and that the collection contains the number $n_i$ $t_i$ times (you can also suppose that the $n_i$ are sorted)
Then your condition is that for every number $x$ between 0 and $N$, $x$ can be written as $x = \Sigma u_i n_i$ with $0 \leq u_i \leq t_i$ in exactly one way.
This is possible if and only if $n_1 = 1$, forall $i$, $n_i = \Sigma_{j\lt i} t_i n_i$, and $N = \Sigma t_i n_i$.
You can prove this by doing an induction on $k$, starting at $k=N=0$ for the base case.
The base case is easy. The induction case is not difficult :
If you have a valid collection containing $k$ distinct terms that can make all numbers up to $N_k$, then the only way to extend it into a bigger collection is by adding $n_{k+1} = N_k+1$.
If you pick a smaller number, then $n_{k+1}$ can be written in two ways, if you pick a bigger number, then you can not make $N_k+1$.
And then, if you have a valid collection made of $k+1$ terms, then the sub-collection containing the first $k$ distinct terms is also valid and has to write every number up to $n_{k+1}-1$ for the same kind of reasons.
Thus, a valid collection of $k$ distinct terms that makes every number up to $N$ is determined by the sequence $(n_1=1, \ldots n_k, n_{k+1}=N+1)$ where $n_i$ divides $n_{i+1}$ :
 $$n_i = \Sigma_{j \lt i} t_j n_j = n_{i-1} +  t_{i-1} n_{i-1} = (t_{i-1} + 1) n_{i-1}$$
So this shows why they are successives multiples of each other and how to recover the $t_i$ from the sequence.
The 4 collections you gave as an example correspond respectively to the sequences $(1,8), (1,2,8), (1,4,8), (1,2,4,8)$.
The number of valid sequences depends on the exponents in the prime decomposition of $N+1$ :
if $N+1$ is a prime power $p^a$, then there are exactly $2^{a-1}$ valid sequences, since you have to choose wether you pick $p^i$ or not for $0 \lt i \lt a$.
If $N+1$ has several prime divisors, I don't think there is a nice formula giving the number of valid collections from the multiset of exponents in the prime factorisation of $N+1$. 
