Example 11 Let $n$ be a given natural number, $n \geq 3$. For $n$ given real numbers $a_{1}, a_{2}, \cdots, a_{n}$, the minimum value of $a_{i}-a_{j}$ $(1 \leqslant i, j \leqslant n)$ is $m$. Find the maximum value of $m$ under the condition $a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}=1$ (1992 Shanghai Mathematics Competition).
$$\begin{array}{l}
a_{1} \geqslant 2 m, \cdots, a_{n}-a_{n-1} \geqslant m, a_{j}-a_{i} \geqslant(j-i) m\left(\mathrm{I}_{-} \leqslant i<j \leqslant n\right) . \\
\sum_{i \leqslant i<j \leqslant u}\left(a_{i}-a_{j}\right)^{2} \geqslant m^{2} \sum_{i \leqslant i<j \leq n}(i-j)^{2}=m^{2} \frac{h(k+1)(2 k+1)}{6}= \\
\frac{m^{2}}{6} \sum_{i=1}^{n}[2 k(k+1)(k+2)-3 k(k+1) j= \\
\frac{m^{2}}{6}\left[\frac{1}{2}(n-1) n(n+1)(n+2)-(n-1) n(n+1)\right]= \\
\frac{m^{2}}{12} n^{2}\left(n^{2}-1\right)
\end{array}$$

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

Therefore,
$$\begin{array}{l}
n \geqslant \frac{m^{2}}{12} n^{2}\left(n^{2}-1\right) \\
m \leqslant \sqrt{\frac{12}{n\left(n^{2}-1\right)}}
\end{array}$$

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