Since $\operatorname{sign}(x+y)$ can take one of 3 values, $z$ can be one of 3 values: 4037,2018, or -1. The same is true of $x$ and $y$. However, this shows that $x+y$ cannot be 0, so $z$ can only be 4037 or -1. The same is true of $x$ and $y$. Now note that, if any two of $x, y, z$ are -1, then the third one must be 4037. Furthermore, if any one of $x, y, z$ is 4037, then the other two must be -1. Thus, the only possibility is to have exactly two of $x, y, z$ be -1 and the third one be 4037. This means that the only remaining triples are $(-1,-1,4037)$ and its permutations. These all work, so there are exactly 3 ordered triples.