21 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 $m$ when
$$a_{1}^{2}+\cdots+a_{n}^{2}=1$$
21. Let $a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{n}$, then $a_{i}-a_{j} \geqslant(i-j) m$. $\sum_{1 \leqslant i<j \leqslant n}\left(a_{i}-a_{j}\right)^{2}=$ $(n-1) \sum_{i=1}^{n} a_{i}^{2}-2 \sum_{1 \leqslant i<j \leqslant n} a_{i} a_{j}=(n-1) \sum_{i=1}^{n} a_{i}^{2}-\left[\left(\sum_{i=1}^{n} a_{i}\right)^{2}-\sum_{i=1}^{n} a_{i}^{2}\right] \leqslant n \sum_{i=1}^{n} a_{i}^{2}=$ n. On the other hand, $\sum_{1 \leq i<j \leqslant n}\left(a_{i}-a_{j}\right)^{2} \geqslant m^{2} \sum_{1 \leq i<j \leq n}(i-j)^{2}=m^{2} \sum_{k=1}^{n-1}(n-k) \cdot k^{2}=$ $\frac{m^{2} n^{2}\left(n^{2}-1\right)}{12}$. Therefore, $n \geqslant \frac{m^{2} n^{2}\left(n^{2}-1\right)}{12}$, which means $m \leqslant \sqrt{\frac{12}{n\left(n^{2}-1\right)}}$. Moreover, the equality holds when $\left|a_{i}\right|$ forms an arithmetic sequence and $\sum_{i=1}^{n} a_{i}=0$. Hence, the maximum value of $m$ is $\sqrt{\frac{12}{n\left(n^{2}-1\right)}}$.