3 Let $n$ be a given positive integer, $n \geqslant 3$, and for $n$ given real numbers $a_{1}, a_{2}, \cdots, a_{n}$, let $m$ be the minimum value of $\left|a_{i}-a_{j}\right|(1 \leqslant i<j \leqslant n)$. Find the maximum value of $m$ under the condition $a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}=1$.
3. Let $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n}$, then
$$\begin{aligned}
a_{j}-a_{i} & =\left(a_{j}-a_{j-1}\right)+\left(a_{j-1}-a_{j-2}\right)+\cdots+\left(a_{i+1}-a_{i}\right) \\
& \geqslant(j-i) m
\end{aligned}$$

Therefore, $\sum_{1 \leqslant i<j \leqslant n}\left(a_{i}-a_{j}\right)^{2} \geqslant \frac{1}{12} n^{2}(n-1)(n+1) m^{2}$,
and since $\sum_{1 \leq i<j \leqslant n}\left(a_{i}-a_{j}\right)^{2}=n\left(a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}\right)-\left(a_{1}+a_{2}+\cdots+a_{n}\right)^{2} \leqslant n$, it follows that $m \leqslant \sqrt{\frac{12}{(n-1) n(n+1)}}$. It is easy to see the condition under which equality holds.