1. Let $A$ be a finite set, for any $x, y \in A$, if $x \neq y$, then $x+y \in A$. Then, the maximum number of elements in $A$ is $\qquad$ .
Let the number of elements in $A$ be $n$.
If $n>3$, then there must be two elements with the same sign. Without loss of generality, assume there are two positive numbers. Let the largest two positive numbers be $a, b$ (with $a < b$). Then, $a+b \notin A$, which is a contradiction.
Therefore, $n \leqslant 3$.
Also, $A=\{0,1,-1\}$ satisfies the condition, so the maximum value of $n$ is 3.