1. Let $x$ and $y$ be real numbers, and satisfy
$$
\left\{\begin{array}{l}
(x-1)^{5}+1999(x-1)=2519, \\
(y-1)^{5}+1999(y-1)=-2519 .
\end{array}\right.
$$

Then the value of $x+y$ is $\qquad$
1.2.
Let $f(t)=t^{5}+1999 t$. Then $f(t)$ is increasing on $R$, and $f(x-1)=-f(y-1)$. Since $f(y-1)=-f(1-y)$, then
$$
\begin{array}{l}
f(x-1)=f(1-y) . \\
\therefore x-1=1-y \Rightarrow x+y=2 .
\end{array}
$$