Cute problem. So note that we can assume that not all three numbers are even, because we could just divide by 4 the whole expression. It's obvious that $x+y+z$ can't be odd because$x+y+z\equiv x^2+y^2+z^2 \equiv 0\pmod{2}$, so we conclude that 2 of them are odd while one of them is even. WLOG let $x$ be even. Add $2(xy+xz+yz)$ to both sides to obtain $$(x+y+z)^2=2018(xy+xz+yz)$$. Now take a look at number 2. We have that $2|ab+bc+ac$ because $V_2(RHS)$ must be even since it is a square. This implies $2|bc$, contradiction. Note that the condition that they are integers doesn't bother us, because $u\equiv -u\pmod{2}$. Also if one of them was 0, the process would be identical. So no triples satisfy the given condition. This problem could be solved with any integer divisible by 3, not only 2016.

Pinko wrote: Find all triples $(x,y,z)$, $x,y,z\in \mathbb{Z}$ for which the number 2016 can be presented as $\frac{x^2+y^2+z^2}{xy+yz+zx}$. $4|x^2+y^2+z^2$ and if we assume as above that not all of them are even, we get a contradiction.

Almost the same as Iran TST 2013 https://artofproblemsolving.com/community/c6h530379p3026714