jmc

Since $\mathrm{sign}(x+y)$ can be $-1,$ 0, or 1, $z$ can be 4037, 2018, or $-1.$ The same holds for $x$ and $y.$ But we can then check that $x+y$ cannot be 0, so $z$ can only be 4037 or $-1.$ And again, the same holds for $x$ and $y.$

If any two of $x,$ $y,$ and $z$ are equal to $-1,$ then the third must be equal to 4037. Conversely, if any of $x,$ $y,$ $z$ are equal to 4037, then the other two must be equal to $-1.$ Therefore, the only solutions are $(4037,-1,-1),$ $(-1,4037,-1),$ and $(-1,-1,4037),$ giving us $\boxed{3}$ solutions.