Example 36 (1992 Shanghai High School Competition Question) 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 that $a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}=$ 1.
We start with the special case of $n=3$, without loss of generality, assume $a_{1} \leqslant a_{2} \leqslant a_{3}$, then $a_{2}-a_{1} \geqslant m, a_{3}-a_{2} \geqslant m, a_{3}-a_{1} \geqslant 2 m$, thus
$$
\begin{aligned}
6 m^{2} & =m^{2}\left(1^{2}+1^{2}+2^{2}\right) \leqslant\left(a_{1}-a_{2}\right)^{2}+\left(a_{2}-a_{3}\right)^{2}+\left(a_{1}-a_{3}\right)^{2} \\
& =2\left(a_{1}^{2}+a_{2}^{2}+a_{3}^{2}\right)-2\left(a_{1} a_{2}+a_{2} a_{3}+a_{1} a_{3}\right) \\
& =3\left(a_{1}^{2}+a_{2}^{2}+a_{3}^{2}\right)-\left(a_{1}+a_{2}+a_{3}\right)^{2} \leqslant 3 .
\end{aligned}
$$

Therefore, $m \leqslant \frac{1}{\sqrt{2}}$, equality holds if and only if $a_{3}-a_{2}=a_{2}-a_{1}$ and $a_{1}+a_{2}+a_{3}=0$, combining with the given condition $a_{1}^{2}+a_{2}^{2}+a_{3}^{2}=1$, equality holds if and only if $a_{1}=-\frac{1}{\sqrt{2}}, a_{2}=0, a_{3}=\frac{1}{\sqrt{2}}$, hence the minimum value of $m$ is $\frac{1}{\sqrt{2}}$.
From the above reasoning, it is not difficult to generalize to the case of $n$ real numbers.
Solution Without loss of generality, assume $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n}$, then for any $1 \leqslant i<j \leqslant n$, we have $a_{j}-a_{i}=\left(a_{j}-\right.$ $\left.a_{j-1}\right)+\left(a_{j-1}-a_{j-2}\right)+\cdots+\left(a_{i+1}-a_{i}\right) \geqslant m+m+\cdots+m=(j-i) m$, thus
$$
\begin{aligned}
m^{2} \sum_{1 \leqslant i<j \leqslant n}(j-i)^{2} & \leqslant \sum_{1 \leqslant i<j \leqslant n}\left|a_{i}-a_{j}\right|^{2} \\
& =(n-1) \sum_{i=1}^{n} a_{i}^{2}-2 \sum_{1 \leqslant i<j \leqslant n} a_{i} a_{j} \\
& =n \sum_{i=1}^{n} a_{i}^{2}-\left(\sum_{i=1}^{n} a_{i}\right)^{2} \\
& \leqslant n .
\end{aligned}
$$

And from (1) we have $\sum_{1 \leqslant i<j \leqslant n}(j-i)^{2}=n \sum_{i=1}^{n} i^{2}-\left(\sum_{i=1}^{n} i\right)^{2}$
$$
\begin{array}{l}
=n \cdot \frac{1}{6} n(n+1)(2 n+1)-\left[\frac{n(n+1)}{2}\right]^{2} \\
=\frac{n^{2}(n+1)(n-1)}{12}=\frac{1}{12} n^{2}\left(n^{2}-1\right),
\end{array}
$$

Therefore, $m \leqslant \sqrt{\frac{12}{n\left(n^{2}-1\right)}}$,
equality holds if and only if $a_{2}-a_{1}=a_{3}-a_{2}=\cdots=a_{n}-a_{n-1}, a_{1}+a_{2}+\cdots+a_{n}=0$, combining with $a_{1}^{2}+$ $a_{2}^{2}+\cdots+a_{n}^{2}=1$ can uniquely determine $a_{1}, a_{2}, \cdots, a_{n}$. Hence, the maximum value of $m$ is $\sqrt{\frac{12}{n\left(n^{2}-1\right)}}$.