9. Real numbers $x, y$ satisfy $\left\{\begin{array}{l}x+\sin y=2008, \\ x+2008 \cos y=2007\end{array}\right.$ $\left(0 \leqslant y \leqslant \frac{\pi}{2}\right)$. Then $x+y=$ $\qquad$
9. $2007+\frac{\pi}{2}$.

Subtracting the two equations gives $\sin y=1+2008 \cos y$. Given $0 \leqslant y \leqslant \frac{\pi}{2}$, we know $1+2008 \cos y \geqslant 1$. Therefore, it can only be that $\sin y=1, \cos y=0$. Hence, $y=\frac{\pi}{2}$. Consequently, $x=2007$.

Therefore, $x+y=2007+\frac{\pi}{2}$.