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 the minimum value of $\left|a_{i}-a_{j}\right|(1 \leqslant$ $i<j \leqslant n)$ as $m$. Find the maximum value of $m$ under the condition $\sum_{i=1}^{n} a_{i}^{2}=1$.
11. Solution:

Without loss of generality, let $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n}$,
$$
\sum_{1 \leqslant i<j \leqslant n}\left(a_{i}-a_{j}\right)^{2}=n-1-2 \sum_{1 \leqslant i<j \leqslant n} a_{i} a_{j}=n-\left(\sum_{i=1}^{n} a_{i}\right)^{2} \leqslant n .
$$

On the other hand, $a_{i+1}-a_{i} \geqslant m, a_{i+2}-a_{i+1} \geqslant m, \cdots, a_{i}-a_{i-1} \geqslant m$, adding these inequalities, we get $a_{j}-a_{i} \geqslant(j-i) m$. Therefore, $\sum_{1<i<j<n}\left(a_{i}-a_{j}\right)^{2} \geqslant m^{2} \sum_{1<i<j<n}(i-j)^{2}$
$$
\begin{array}{l}
=m^{2} \sum_{k=1}^{n-1}\left(1^{2}+2^{2}+\cdots+k^{2}\right)=m^{2} \sum_{k=1}^{n-1} \frac{k(k+1)(2 k+1)}{6} \\
=\frac{m^{2}}{6} \sum_{k=1}^{n-1}[2 k(k+1)(k+2)-3 k(k+1)] \\
=\frac{m^{2}}{6}\left[\frac{1}{2}(n-1) n(n+1)(n+2)-(n-1) n(n+1)\right] \\
=\frac{m^{2}}{12} n^{2}(n+1)(n-1),
\end{array}
$$

Thus, $n \geqslant \frac{m^{2}}{12} n^{2}(n+1)(n-1)$,
$m \leqslant \sqrt{\frac{12}{n\left(n^{2}-1\right)}}$, equality holds if and only if $\sum_{i=1}^{n} a_{i}=0$, and $\left\{a_{i}\right\}$ forms an arithmetic sequence.
Therefore, the maximum value of $m$ is $\sqrt{\frac{12}{n\left(n^{2}-1\right)}}$.