I claim that $7$ works. First, to see that $6$ is not enough, see $1,1,1,1,-2$. Taking any two of the $1$s and the $-2$ yields a sum of $0$, and there are $6$ such sums. It just remains to prove that $7$ is enough. Observe that among all $7$ sums, there are $21$ terms, while there are only $5$ numbers, so by Pigeonhole Principle we have some number that appears $5$ times. WLOG $x_1$ appears $5$ times. It is not included in at most one sum, and due to the symmetry, we can suppose $x_1+x_4+x_5$ is not necessarily $0$. Thus $x_1+x_2+x_3 = x_1+x_2+x_4 = x_1+x_2+x_5 = x_1+x_3+x_4 = x_1+x_3+x_5 = 0$. The first three equations simplify to $x_3 = x_4 = x_5$, and the third and fifth equations give $x_2 = x_3$, so $x_2 = x_3 = x_4 = x_5$. Also note that $x_1$ can only participate in at most $6$ sums, but we have $7$ sums that give $0$, so there exists some sum of $x_2,x_3,x_4,x_5$ that adds up to $0$. WLOG $x_2 + x_3 + x_4 = 0$. But $x_2 = x_3 = x_4$, so $x_2 = x_3 = x_4 = 0$, and easily $x_1 = x_5 = 0$. By the way, this is Combinatorics.