9 Real numbers $x, y$ satisfy $\left\{\begin{array}{l}x+\sin y=2008, \\ x+2008 \cos y=2007,\end{array}\right.$ where $0 \leqslant y \leqslant \frac{\pi}{2}$, then $x+y=$
Subtract the two equations, we get
$$
\sin y=1+2008 \cos y .
$$

Given $0 \leqslant y \leqslant \frac{\pi}{2}$, we know
$$
1+2008 \cos y \geqslant 1,
$$

thus it can only be $\sin y=1, \cos y=0$. Hence $y=\frac{\pi}{2}$, and therefore
$$
x=2007, x+y=2007+\frac{\pi}{2} .
$$