Let $2017 = 6k+1$. Add \begin{align*} f(\pm (2k-0)) +f(\pm (4k+1)) + f(\mp (6k+1))&\geq 0\\ f(\pm (2k-2)) +f(\pm (4k+2)) + f(\mp (6k+0))&\geq 0\\ f(\pm (2k-4)) +f(\pm (4k+3)) + f(\mp (6k-1))&\geq 0\\ \cdots\\ f(\pm 2) +f(\pm(5k)) +f(\mp (5k+2))&\geq 0\\ f(\pm(2k-1))+f(\pm(2k+1))+f(\mp(4k-0))&\geq 0\\ f(\pm(2k-3))+f(\pm(2k+2))+f(\mp(4k-1))&\geq 0\\ \cdots\\ f(\pm 1)+f(\pm 3k)+f(\mp(3k+1))&\geq 0\\ f(5k+1) + f(0) + f(-5k-1)&\geq 0 \end{align*}to get the result.

Meh. Tastes like ISL 2011 A5.

drkim wrote: Let $f : \mathbb{Z} \to \mathbb{R}$ be a function satisfying $f(x) + f(y) + f(z) \ge 0$ for all integers $x, y, z$ with $x + y + z = 0$. Prove that \[ f(-2017) + f(-2016) + \cdots + f(2016) + f(2017) \ge 0. \] Nice problem.