Answer is $n=6$, let the numbers $a_1=1,a_2=5,a_3=4,a_4=-1,a_5=-5,a_6=-4$ be in a hexagon in that order. Consider a graph on $n$ vertices $a_i$ and join $a_i$ to $a_j$ and $a_k$ iff $a_i = a_j + a_k$. By the handshake lemma the graph has $n$ edges. Clearly there cannot be a triangle, and it is also easy to show that $n=4$ doesn't work which means a cycle of $4$ cannot exist. Thus we need to rule out $n=5$. Let $a_1,\dots , a_5$ be the numbers in the cycle $a_i = a_{i-1} = a_{i+1}$, adding $a_1 = a_5 + a_2$ and $a_5 = a_1 + a_4$ gives $a_2 + a_4=0$ and adding $a_2 = a_1 + a_3$ and $a_3 = a_2 + a_4$ gives $a_1 + a_4=0\implies a_1 = a_2$ which is not possible.