Example 8 Let $n$ be a given natural number, and $n \geqslant 3$, for $n$ given real numbers $a_{1}, a_{2}, \cdots, a_{n}$, denote $\left|a_{i}-a_{j}\right|$ $(1 \leqslant i<j \leqslant n)$ the minimum value as $m$. Find the maximum value of the above $m$ under the condition $a_{1}^{2}+a_{2}^{2}+$ $\cdots+a_{n}^{2}=1$.
Solution: Without loss of generality, let $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n}$, then
$$
\begin{array}{l} 
a_{2}-a_{1} \geqslant m, a_{3}-a_{2} \geqslant m, \cdots, a_{n}-a_{n-1} \\
\geqslant m, a_{1}-a_{i} \geqslant(j-i) m \quad(1 \leqslant i<j \leqslant n) . \\
\sum_{1 \leqslant i<j \leqslant n}\left(a_{i}-a_{j}\right)^{2} \geqslant m^{2} \sum_{1 \leqslant i<j \leqslant n}(i-j)^{2} \\
= m^{2} \sum_{k=1}^{n-1} \frac{k(k+1)(2 k+1)}{6} \\
= \frac{m^{2}}{6} \sum_{k=1}^{n-1}[2 k(k+1)(k+2)-3 k(k+1)] \\
= \frac{m^{2}}{6}\left[\frac{1}{2}(n-1) n(n+1)(n+2)-\right. \\
(n-1) n(n+1)] \\
= \frac{m^{2}}{12} n^{2}\left(n^{2}-1\right) .
\end{array}
$$

On the other hand, by $a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}=1$, we get
$$
\begin{array}{l}
\sum_{1 \leqslant i<j \leqslant n}\left(a_{i}-a_{j}\right)^{2}=n-1-2 \sum_{1 \leqslant i<j \leqslant n} a_{i} a_{j} \\
=n-\left(\sum_{i=1}^{n} a_{i}\right)^{2} \leqslant n . \\
\text { Hence } \frac{m^{2}}{12} n^{2}\left(n^{2}-1\right) \leqslant n,
\end{array}
$$

i.e., $m \leqslant \sqrt{\frac{12}{n\left(n^{2}-1\right)}}$.
Equality holds if and only if $\sum_{i=1}^{n} a_{i}=0$ and $a_{1}, a_{2}, \cdots, a_{n}$ form an arithmetic sequence, so the maximum value of $m$ is
$$
\sqrt{\frac{12}{n\left(n^{2}-1\right)}} .
$$