11. Given a convex $n$-sided polygon where the degrees of the $n$ interior angles are all integers and distinct, and the largest interior angle is three times the smallest interior angle, the maximum value that $n$ can take is $\qquad$ .
For a convex $n$-sided polygon with $n$ interior angles $\alpha=x_{1}<x_{2}<\cdots<x_{n}=3 \alpha$, then
$$
\begin{array}{l}
(n-2) \cdot 180=x_{1}+x_{2}+\cdots+x_{n} \leqslant 3 \alpha+3 \alpha-1+3 \alpha-2+\cdots+3 \alpha-n+2+\alpha \\
=(3 n-2) \alpha-\frac{(n-1)(n-2)}{2} \Rightarrow n^{2}+(357-6 \alpha) n+4 \alpha-718 \leqslant 0 .
\end{array}
$$

When $\alpha=59$, $n^{2}+3 n-482 \leqslant 0 \Rightarrow n \leqslant 20$;
When $\alpha=58$, $n^{2}+9 n-486 \leqslant 0 \Rightarrow n \leqslant 18$. Therefore, the maximum value of $n$ is 20.