$$
\begin{array}{l}
\text { 9. In } \triangle A B C, \angle A<\angle B<\angle C, \\
\frac{\sin A+\sin B+\sin C}{\cos A+\cos B+\cos C}=\sqrt{3} .
\end{array}
$$

Then $\angle B=$
9. $\frac{\pi}{3}$.

From the given information, we have
$$
\begin{array}{l}
(\sin A - \sqrt{3} \cos A) + (\sin B - \sqrt{3} \cos B) + \\
(\sin C - \sqrt{3} \cos C) = 0 \\
\Rightarrow \sin \alpha + \sin \beta + \sin \gamma = 0,
\end{array}
$$

where $\alpha = \angle A - \frac{\pi}{3}, \beta = \angle B - \frac{\pi}{3}, \gamma = \angle C - \frac{\pi}{3}$.
Then $2 \sin \frac{\alpha + \beta}{2} \cdot \cos \frac{\alpha - \beta}{2} - 2 \sin \frac{\alpha + \beta}{2} \cdot \cos \frac{\alpha + \beta}{2} = 0$
$\Rightarrow 4 \sin \frac{\alpha}{2} \cdot \sin \frac{\beta}{2} \cdot \sin \frac{\gamma}{2} = 0$.
Thus, at least one of $\angle A, \angle B, \angle C$ is $\frac{\pi}{3}$.
Given $\angle A < \angle B < \angle C$, therefore, $\angle B = \frac{\pi}{3}$.