11. Let $n$ be a given natural number, $n \geqslant 3$, and for $n$ given real numbers $a_{1}, a_{2}$, $\cdots, a_{n}$, denote $\left|a_{i}-a_{j}\right|(1 \leqslant i<j \leqslant n)$ by $m$, the minimum value. Find the maximum value of $m$ under the condition $a_{1}^{2}+a_{2}^{2}+\cdots+$ $a_{n}^{2}=1$.
11. Let $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n}$, then $a_{2}-a_{1} \geqslant m, a_{3}-a_{2} \geqslant m, \cdots, a_{n}-a_{n-1} \geqslant m, a_{j}-a_{i} \geqslant(j-i) m(1 \leqslant i<j \leqslant n)$,
$$
\sum_{1 \leqslant i<j \leqslant n}\left(a_{i}-a_{j}\right)^{2} \geqslant m^{2} \sum_{1 \leqslant i<j \leqslant n}(i-j)^{2}=m^{2} \sum_{k=1}^{n-1} \frac{k(k+1)(2 k+1)}{1}=\frac{m^{2}}{12} n^{2}\left(n^{2}-1\right) .
$$

On the other hand, by $a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}=1$, we get $\sum_{1 \leqslant i<j \leqslant n}\left(a_{i}-a_{j}\right)^{2}=n-\left(\sum_{i=1}^{n} a_{i}\right)^{2} \leqslant n$. Therefore, $n \geqslant \frac{m^{2}}{12} n^{2}\left(n^{2}-1\right)$. Hence, $m \leqslant \sqrt{\frac{12}{n\left(n^{2}-1\right)}}$, with equality if and only if $\sum_{i=1}^{n} a_{i}=0$ and $a_{1}, a_{2}, \cdots, a_{n}$ form an arithmetic sequence.