The answer is 7, achieved when $M=\{-3, -2, -1, 0, 1, 2, 3 \}$ We will prove that every working $M$ has at most $3$ elements with the same sign. Suppose otherwise; let $a>b>c>d>0$ be the largest elements of $M$. It's clear that $(a,b,c)$ and $(a,b,d)$ cannot both satisfy the condition (size reasoning). A similar argument holds for negatives, which means that $M$ has at most $6$ nonzero elements. Done.