, 2012

Solution:
The set $A=\{F_{2 n}: n=1,2, \ldots\} \cup\{-F_{2 n+1}: n=1,2, \ldots\}$, where $F_{k}$ is the $k^{\text{th}}$ Fibonacci number ($F_{1}=1$, $F_{2}=1$, $F_{k+2}=F_{k+1}+F_{k}$ for $k \geq 1$) qualifies for an example. We then have $F_{2 n}=F_{2 n+2}+(-F_{2 n+1})$ and $-F_{2 n+1}=(-F_{2 n+3})+F_{2 n+2}$ for all $n \geq 1$, so $A$ is sum-full (and even with unique representations). On the other hand, we can never have
$$0=\sum_{i=1}^{s} F_{2 n_{i}}-\sum_{j=1}^{t} F_{2 n_{j}+1}$$
owing to the fact that Zeckendorf representations are known to be unique.

It remains to be shown that all nonzero values can be represented as sums of distinct numbers $1,-2,3,-5,8,-13,21, \ldots$. This may be done using a greedy algorithm: when representing $n$, the number largest in magnitude that is used is the element $m= \pm F_{k}$ of $A$ that is closest to $0$ subject to having the same sign as $n$ and $|m| \geq |n|$. That this algorithm terminates without using any member of $A$ twice is a straightforward induction on $k$; the base case is $k=2$ ($m=1$) and the induction hypothesis is that for all $n$ for which the above algorithm starts with $\pm F_{\ell}$ with $\ell \leq k$, it terminates without having used any member of $A$ twice and without having used any $\pm F_{j}$ with $j>\ell$.

Remark (James Aaronson and Adam P Goucher). Let $n$ be a positive integer, and write $u=2^{n}$; we claim that the set
$$\{1,2,4, \ldots, 2^{n-1},-u, u+1,-(2 u+1), 3 u+2,-(5 u+3), 8 u+5, \ldots\}$$
is a sum-full zero-sum-free set. The proof is similar to that used for the standard examples.