Example 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 the above $m$ under the condition $a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}=1$. (1992 Shanghai Mathematics Competition Problem)
Let's assume $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n}$. Then $a_{2}-a_{1} \geqslant m, a_{3}-a_{2} \geqslant m$,
$$\begin{aligned}
a_{3}-a_{1} & \geqslant 2 m, \cdots, a_{n}-a_{n-1} \geqslant m, a_{j}-a_{i} \geqslant(j-i) m(1 \leqslant i<j \leqslant n) \\
& \sum_{1 \leqslant i<j \leqslant n}\left(a_{i}-a_{j}\right)^{2} \geqslant m^{2} \sum_{1 \leqslant i<j \leqslant n}(i-j)^{2} \\
= & m^{2} \sum_{k=1}^{n-1} \frac{k(k+1)(2 k+1)}{6} \\
= & \frac{m^{2}}{6} \sum_{k=1}^{n-1}[2 k(k+1)(k+2)-3 k(k+1)] \\
= & \frac{m^{2}}{6}\left[\frac{1}{2}(n-1) n(n+1)(n+2)-(n-1) n(n+1)\right] \\
= & \frac{m^{2}}{12} n^{2}\left(n^{2}-1\right)
\end{aligned}$$

On the other hand, from $a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}=1$ we get
$$\sum_{1 \leqslant j<j \leqslant n}\left(a_{i}-a_{j}\right)^{2}=n-1-2 \sum_{1 \leqslant i<j \leqslant n} a_{i} a_{j}=n-\left(\sum_{i=1}^{n} a_{i}\right)^{2} \leqslant n,$$

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