7. $[4]$ For any real number $\alpha$, define
$$
\operatorname{sign}(\alpha)=\left\{\begin{array}{ll}
+1 & \text { if } \alpha>0 \\
0 & \text { if } \alpha=0 \\
-1 & \text { if } \alpha<0
\end{array}\right.
$$

How many triples $(x, y, z) \in \mathbb{R}^{3}$ satisfy the following system of equations
$$
\begin{array}{l}
x=2018-2019 \cdot \operatorname{sign}(y+z) \\
y=2018-2019 \cdot \operatorname{sign}(z+x) \\
z=2018-2019 \cdot \operatorname{sign}(x+y) ?
\end{array}
$$
Answer: $\square$
Since 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.