1. Two different numbers $x$ and $y$ (not necessarily integers) are such that $x^{2}-2000 x=y^{2}-2000 y$. Find the sum of the numbers $x$ and $y$.
1. Move $2000 x$ to the right side, and $y^{2}$ - to the left. We get:

$$
x^{2}-y^{2}=2000 x-2000 y
$$

Factor the left side using the difference of squares formula:

$$
(x-y)(x+y)=2000(x-y) .
$$

Since $x \neq y$, we can cancel out $(x-y)$. Thus, $x+y=$ $=2000$.