Example 26 Let $n(\geqslant 3)$ be a given positive integer, 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 that $a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}=$ 1.
Let's assume $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n}$. We estimate $\sum_{1 \leqslant i<j \leqslant n}\left(a_{i}-a_{j}\right)^{2}$ from two aspects.

On one hand,
$$
\begin{array}{c}
\sum_{1 \leqslant i=j \leqslant i}\left(a_{i}-a_{j}\right)^{2}=(n-1) \sum_{i=1}^{n} a_{i}^{2}-2 \sum_{1{ }_{1} i<j \leqslant n} a_{i} a_{j} \\
=n \sum_{i=1}^{n} a_{i}^{2}-\left(\sum_{i=1}^{n} a_{i}\right)^{2} \leqslant n .
\end{array}
$$

On the other hand, since $a_{2}-a_{1} \geqslant m, a_{3}-a_{2} \geqslant m, \cdots, a_{n}-a_{n-1} \geqslant m$, for $1 \leqslant i<j \leqslant n$, we have $a_{j}-a_{i} \geqslant(j-i) m$.
Thus,
$$
\begin{array}{l}
\sum_{1 \leqslant i<j \leqslant n}\left(a_{i}-a_{j}\right)^{2} \geqslant m^{2} \sum_{1 \leqslant i<j \leqslant n}(i-j)^{2} \\
=m^{2} \sum_{k=1}^{n-1}(n-k) k^{2}=\frac{m^{2}}{12} n^{2}\left(n^{2}-1\right) .
\end{array}
$$

From the above two aspects, we know

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

Furthermore, when $\sum_{i=1}^{n} a_{i}=0$, and $a_{1}, a_{2}, \cdots, a_{n}$ form an arithmetic sequence, the above inequality holds with equality. Hence, the maximum value of $m$ is $\sqrt{\frac{12}{n\left(n^{2}-1\right)}}$.