For every finite set of real numbers, $S =\{a_1,a_2,...,a_n\}$ which satisfies the condition of the problem, there corresponds a directed graph $G_S$ with the following construction:

Each element of $S$ is associated with a vertex in $G_S$, according to the rule that if $a_i \in S $ then $ V_i \in G_S$. A vertex $V_i \in G_S$ is connected to another vertex $V_k \in G_S$ by an edge $E_j$ if and only if $a_i + a_j = a_k$. Clearly, the existence of a closed cycle in $G_s$ with no repeat edges corresponds to a zero sum subset of $S$.

A useful fact, which I refer to as the "vertex replacement trick", is that if $V_i \rightarrow(E_j)\rightarrow V_k$, then by the defining property of $S$, we also have $V_j \rightarrow(E_i)\rightarrow V_k$.

I will say that a path $P$ is "well behaved" if no index of $S$ appears more than once in $P$, unless it occurs in an adjacent vertex/edge pair of the form $ \cdots \rightarrow V_i \rightarrow (E_i) \rightarrow \cdots$

Choose an arbitrary vertex $V_a \in G_S$, and choose an arbitrary incoming edge to $V_a$, say $E_b$. Then the initial path is:

$$P=V_c \rightarrow (E_b) \rightarrow V_a $$

Unless $0 \in S$, then every possible initial path is well behaved.

Add one vertex/edge pair to $P$, if $P$ is still well behaved then repeat until it is not. Due to the defining property of $S$, it's always possible to find another vertex to add to any given path, and since $S$ is finite, this process must eventually terminate.

$Claim:$ If $P$ is well behaved, but $P'=V_y \rightarrow (E_x) \rightarrow P$ is not, then $P'$ either contains a closed cycle with no repeat edges, or it can be transformed into one that does by applying the vertex replacement trick.

$Proof:$ If the index $x$ does not appear in $P$ but $y$ does, then we can use the vertex replacement trick to ensure that $V_y \in P$, and thus obtain a closed cycle, which by virtue of $P$ being well behaved, is guaranteed to have no repeat edges:

$$V_y \rightarrow E_x \rightarrow \cdots \rightarrow V_y$$

If the index $y$ does not appear in $P$ but $x$ does, then similarly, we use the vertex replacement trick to obtain:

$$V_x \rightarrow E_y \rightarrow \cdots \rightarrow V_x$$

In the case that $x=y$, we obtain:

$$V_x \rightarrow E_x \rightarrow \cdots \rightarrow V_x$$

Otherwise, in the case that the indices $x$ and $y$ both appear in $P$, choose the element with index of $x$ or $y$ which occurs first in $P$, and if that element is an edge, turn it into a vertex using the trick. Then remove all the elements of $P$ which occur after that vertex, and return to one of the previous cases where only one of the indices appears in $P$.

EDIT: This is a massive update/rewrite of an unfinished answer of mine from years ago. My apologies if this makes the old comments no longer relevant.

`$a_i = a_j + a_k$`

restrict the vector`$(a_1, \dots, a_n)$`

to a linear subspace $L$ of $\mathbb{R}^n$. There are only $2^n-1$ conditions that a sum be 0, so each one is a hyperplane. If there is no solution then each such hyperplane intersected with $L$ is a hyperplane in that subspace. But the finite union of hyperplanes can't be the whole space, so there is some point with rational coordinates (since a hyperplane is closed -- the remaining set is open). $\endgroup$ – Victor Miller Mar 4 '10 at 14:29