Let $m = 2n+1$ and make a directed graph $G$ on $\mathbb{Z}/m\mathbb{Z}$ where $u \mapsto v$ if and only if there exists some $a_j$ congruent to $(v-u) \bmod m$. Note that every vertex in $G$ has outdegree exactly $n$ now, so there are $n(2n+1) = {m \choose 2}$ edges. Claim: $G$ is a tournament. Proof: Observe that if $k \in \mathbb{Z}/m\mathbb{Z}$ is present among the $a_i$, then $-k$ cannot be. As a result, if $u \mapsto v$ then we cannot possibly have $v \mapsto u$. So there is at most one edge between any pair of vertices in either direction. Since there are exactly ${m \choose 2}$ edges, $G$ is a tournament. Claim: $G$ has a cycle containing exactly $n$ vertices. Proof: Since $G$ is a tournament, by the infamous Twitch Lemma it suffices to show that $G$ has a cycle with at least $n$ vertices. Let \[v_1 \mapsto v_2 \mapsto \dots \mapsto v_{\ell}\]denote the longest path in $G$. Note that since the path cannot be extended any further, all $n$ edges coming out of $v_{\ell}$ lead to some vertex among $v_1, v_2, \dots, v_{\ell-1}$. Consider the edge pointing to the $v_i$ with $i$ minimum. Then $i \leq \ell-n$ and thus the cycle \[v_{i} \mapsto v_{i+1} \mapsto \dots \mapsto v_{\ell} \mapsto v_{i}\]has $\ell-i+1 \geq n+1$ vertices, proving the claim. Suppose the vertices of the $n$-cycle are $v_1, v_2, \dots, v_n$. For $1 \leq i \leq n$ let $k_i$ denote the index such that $a_{k_i} \equiv v_{i+1}-v_i \pmod m$ where the indices are read modulo $n$. It is now easy to see that these $k_i$ fit the requirements of the condition, since the partial sums above are, in that order \[v_2-v_1, v_3-v_1, \dots, v_{n}-v_1, v_1-v_1\](that is, they all are distinct modulo $m$ and the last sum is divisible by $m$) This concludes the solution. $\square$