Example 3 Given the $n$ interior angles of an $n$-sided polygon, sorted in descending order as $A_{1}>A_{2}>\cdots>A_{n}(n \geqslant 4)$, let $\alpha$ denote the minimum of $180^{\circ}-A_{1}, A_{1}-A_{2}, \cdots, A_{n-1}-A_{n}$. Find the maximum value of $\alpha$.

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The translation maintains the original text's line breaks and formatting.
$$
\begin{array}{l}
\text { Solution: } \alpha=\frac{n \alpha+(n-1) \alpha+\cdots+\alpha}{n+(n-1)+\cdots+1} \\
\leqslant \frac{n\left(180^{\circ}-A_{1}\right)+(n-1)\left(A_{1}-A_{2}\right)+\cdots+\left(A_{n-1}-A_{n}\right)}{n+(n-1)+\cdots+1} \\
=\frac{2}{n(n+1)}\left[n \times 180^{\circ}-\left(A_{1}+A_{2}+\cdots+A_{n}\right)\right] \\
=\frac{2}{n(n+1)}\left[n \times 180^{\circ}-(n-2) \times 180^{\circ}\right] \\
=\frac{720^{\circ}}{n(n+1)} .
\end{array}
$$

On the other hand, for $\alpha$ to reach $\frac{720^{\circ}}{n(n+1)}$, it is necessary and sufficient that $180^{\circ}-A_{1}=A_{1}-A_{2}=\cdots=A_{n-1}-A_{n}=$ $\frac{720^{\circ}}{n(n+1)}$, that is, $A_{0}=180^{\circ}, A_{1}, A_{2}, \cdots, A_{n}$ form an arithmetic sequence with the first term $180^{\circ}$ and common difference $d=-\frac{720^{\circ}}{n(n+1)}$. Therefore, we have
$$
\begin{array}{l}
A_{k}=A_{0}+k\left[-\frac{720^{\circ}}{n(n+1)}\right] \\
=\frac{n(n+1)-4 k}{n(n+1)} \cdot 180^{\circ}(n \geqslant 4) .
\end{array}
$$

Thus, when the interior angles of an $n$-sided polygon are
$$
A_{k}=\frac{n(n+1)-4 k}{n(n+1)} \cdot 180^{\circ}
$$
$k=1,2, \cdots, n$, $\alpha$ reaches its maximum value $\frac{720^{\circ}}{n(n+1)}$.