7. Let $x, y \in \mathbf{R}$, and satisfy
$$
\left\{\begin{array}{l}
(x-1)^{2003}+2002(x-1)=-1, \\
(y-2)^{2008}+2002(y-2)=1 .
\end{array}\right.
$$

Then $x+y=$
7. 3
Construct the function $f(t)=t^{2008}+2002 t$. It is easy to see that $f(t)$ is an odd function on $\mathbf{R}$, and it is also a monotonically increasing function. From this, we can get $f(x-1)=-f(y-2)$, which means $f(x-1)=f(2-y)$. Therefore, $x-1=2-y, x+y=3$.