9. In $\triangle A B C$, $\angle A<\angle B<\angle C$, $\frac{\sin A+\sin B+\sin C}{\cos A+\cos B+\cos C}=\sqrt{3}$.
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 $2 \sin \frac{\alpha+\beta}{2} \cdot \cos \frac{\alpha-\beta}{2}-2 \sin \frac{\alpha+\beta}{2} \cdot \cos \frac{\alpha+\beta}{2}=0$
$$
\Rightarrow 4 \sin \frac{\alpha}{2} \cdot \sin \frac{\beta}{2} \cdot \sin \frac{\gamma}{2}=0.
$$

Therefore, at least one of $\angle A, \angle B, \angle C$ is $\frac{\pi}{3}$.
Given $\angle A<\angle B<\angle C$, thus, $\angle B=\frac{\pi}{3}$.