A zero sum subset of a sum-full set I had seen this problem a long time back and wasn't able to solve it. For some reason I was reminded of it and thought it might be interesting to the visitors here.

Apparently, this problem is from a mathematics magazine of some university in the United States (sorry, no idea about either).

So the problem is:

Suppose $S \subset \mathbb{Z}$ (set of integers) such that 

1) $|S| = 15$

2) $\forall ~s \in S, \exists ~a,b \in S$ such that $s = a+b$

Show that for every such $S$, there is a non-empty subset $T$ of $S$ such that the sum of elements of $T$ is zero and $|T| \leq 7$.

Update (Sep 13)

Here is an approach which seems promising and others might be able to take it ahead perhaps.

If you look at the set as a vector $s$, then there is a matrix $A$ with the main diagonal being all $1$, each row containing exactly one $1$ and one $-1$ (or a single $2$) in the non-diagonal position such that $As = 0$.

The problem becomes equivalent to proving that for any such matrix $A$ the row space of $A$ contains a vector with all zeroes except for a $1$ and $-1$ or a vector with all zeroes except $\leq 7$ ones.

This implies that the numbers in the set $S$ themselves don't matter and we can perhaps replace them with elements from a different field (like say reals, or complex numbers).
A weaker statement, where we allow elements in $T$ to be repeated, can be proved as below: 

Since we can look at the set $\{-s | s\in S \}$ we may assume there are at most $7$ positive numbers in $S$. Let each positive number be a vertex, from each vertex $s$ we draw an arrow to any vertex $a$ such that $s=a+b$. Since if $s>0$, one of $a,b$ must be positive, there is at least one arrow from any vertex. So there must be a cycle $s_1,\cdots,s_n=s_1$ with $n\leq 8$. We can let $T$ consists of $s_i-s_{i+1}$, $1\leq i\leq n-1$.
In computer science, the subset sum problem is an important decision problem in complexity theory and cryptography. There are several equivalent formulations of the problem. One of them is: given a set (or multiset) of integers, is there a non-empty subset whose sum is zero? For example, given the set {\displaystyle {-7,-3,-2,5,8}} {\displaystyle {-7,-3,-2,5,8}}, the answer is yes because the subset {\displaystyle {-3,-2,5}} {\displaystyle {-3,-2,5}} sums to zero. The problem is NP-complete, meaning roughly that while it is easy to confirm whether a proposed solution is valid, it may inherently be prohibitively difficult to determine in the first place whether any solution exists.

The problem can be equivalently formulated as: given the integers or natural numbers {\displaystyle w_{1},\ldots ,w_{n}} w_{1},\ldots ,w_{n} does any subset of them sum to precisely {\displaystyle W} W?[1] Subset sum can also be thought of as a special case of the knapsack problem.[2] One interesting special case of subset sum is the partition problem, in which s is half of the sum of all elements in the set.

The complexity of the subset sum problem can be viewed as depending on two parameters, N, the number of decision variables, and P, the precision of the problem (stated as the number of binary place values that it takes to state the problem). (Note: here the letters N and P mean something different from what they mean in the NP class of problems.)

The complexity of the best known algorithms is exponential in the smaller of the two parameters N and P. Thus, the problem is most difficult if N and P are of the same order. It only becomes easy if either N or P becomes very small.

If N (the number of variables) is small, then an exhaustive search for the solution is practical. If P (the number of place values) is a small fixed number, then there are dynamic programming algorithms that can solve it exactly.

Efficient algorithms for both small N and small P cases are given below.

There are several ways to solve subset sum in time exponential in {\displaystyle n} n. The most naïve algorithm would be to cycle through all subsets of {\displaystyle n} n numbers and, for every one of them, check if the subset sums to the right number. The running time is of order {\displaystyle O(2^{n}n)} O(2^{n}n), since there are {\displaystyle 2^{n}} 2^{n} subsets and, to check each subset, we need to sum at most {\displaystyle n} n elements.

A better exponential time algorithm is known which runs in time O(2N/2). The algorithm splits arbitrarily the N elements into two sets of N/2 each. For each of these two sets, it stores a list of the sums of all 2N/2 possible subsets of its elements. Each of these two lists is then sorted. Using a standard comparison sorting algorithm for this step would take time O(2N/2N). However, given a sorted list of sums for k elements, the list can be expanded to two sorted lists with the introduction of a (k + 1)st element, and these two sorted lists can be merged in time O(2k). Thus, each list can be generated in sorted form in time O(2N/2). Given the two sorted lists, the algorithm can check if an element of the first array and an element of the second array sum up to s in time O(2N/2). To do that, the algorithm passes through the first array in decreasing order (starting at the largest element) and the second array in increasing order (starting at the smallest element). Whenever the sum of the current element in the first array and the current element in the second array is more than s, the algorithm moves to the next element in the first array. If it is less than s, the algorithm moves to the next element in the second array. If two elements with sum s are found, it stops. Horowitz and Sahni first published this algorithm in a technical report in 1974.[3]

The problem can be solved in pseudo-polynomial time using dynamic programming. Suppose the sequence is

x1, ..., xN
sorted in the increasing order and we wish to determine if there is a nonempty subset which sums to zero. Define the boolean-valued function Q(i, s) to be the value (true or false) of

"there is a nonempty subset of x1, ..., xi which sums to s".
Thus, the solution to the problem "Given a set of integers, is there a non-empty subset whose sum is zero?" is the value of Q(N, 0).

Let A be the sum of the negative values and B the sum of the positive values. Clearly, Q(i, s) = false, if s &lt; A or s > B. So these values do not need to be stored or computed.

Create an array to hold the values Q(i, s) for 1 ≤ i ≤ N and A ≤ s ≤ B.

The array can now be filled in using a simple recursion. Initially, for A ≤ s ≤ B, set

Q(1, s) := (x1 == s)
where == is a boolean function that returns true if x1 is equal to s, false otherwise.

Then, for i = 2, …, N, set

Q(i, s) := Q(i − 1, s) or (xi == s) or Q(i − 1, s − xi),  for A ≤ s ≤ B.
For each assignment, the values of Q on the right side are already known, either because they were stored in the table for the previous value of i or because Q(i − 1,s − xi) = false if s − xi &lt; A or s − xi > B. Therefore, the total number of arithmetic operations is O(N(B − A)). For example, if all the values are O(Nk) for some k, then the time required is O(Nk+2).

This algorithm is easily modified to return the subset with sum 0 if there is one.

The dynamic programming solution has runtime of {\displaystyle O(sN)} O(sN) where {\displaystyle s} s is the sum we want to find in set of {\displaystyle N} N numbers. This solution does not count as polynomial time in complexity theory because B − A is not polynomial in the size of the problem, which is the number of bits used to represent it. This algorithm is polynomial in the values of A and B, which are exponential in their numbers of bits.

For the case that each xi is positive and bounded by a fixed constant C, Pisinger found a linear time algorithm having time complexity O(NC) (note that this is for the version of the problem where the target sum is not necessarily zero, otherwise the problem would be trivial).[4] In 2015, Koiliaris and Xu found the {\displaystyle {\tilde {O}}(s{\sqrt {N}})} {\tilde {O}}(s{\sqrt {N}}) algorithm for the subset sum problem where {\displaystyle s} s is the sum we need to find.[5]