Example 2 Let $n \geqslant 2$,
$$
A_{n}=\left\{x \in \mathbf{R} \left\lvert\, x=\left[\frac{x}{2}\right]+\left[\frac{x}{3}\right]+\cdots+\left[\frac{x}{n}\right]\right.\right\} .
$$
Prove: $A=\underset{n \geqslant 2}{\cup} A_{n}$ is a finite set, and find the maximum and minimum elements of $A$.
(2010, Romanian Mathematical Olympiad (Final)) [Analysis] Start with simple cases in mathematical experiments.
Elements in $A_{2}$ satisfy $x=\left[\frac{x}{2}\right]$, and $x \in \mathbf{Z}$. Hence, $\frac{x-1}{2} \leqslant x=\left[\frac{x}{2}\right] \leqslant \frac{x}{2}$, which means $-1 \leqslant x \leqslant 0$. After verification, $x=0,-1$ are elements of $A_{2}$. Therefore, $A_{2}=\{-1,0\}$.
Elements in $A_{3}$ satisfy
$$
\begin{array}{l}
x=\left[\frac{x}{2}\right]+\left[\frac{x}{3}\right] \text {, and } x \in \mathbf{Z} . \\
\text { Hence } \frac{x-1}{2}+\frac{x-2}{3} \leqslant x=\left[\frac{x}{2}\right]+\left[\frac{x}{3}\right] \\
\leqslant \frac{x}{2}+\frac{x}{3},
\end{array}
$$

which means $-7 \leqslant x \leqslant 0$.
After verification, $x=-7,-5,-4,-3,-2,0$ are elements of $A_{3}$, i.e.,
$$
A_{3}=\{-7,-5,-4,-3,-2,0\} \text {. }
$$
Elements in $A_{4}$ satisfy
$$
\begin{array}{l}
x=\left[\frac{x}{2}\right]+\left[\frac{x}{3}\right]+\left[\frac{x}{4}\right], \text { and } x \in \mathbf{Z} . \\
\text { Hence } \frac{x-1}{2}+\frac{x-2}{3}+\frac{x-3}{4} \leqslant x \\
=\left[\frac{x}{2}\right]+\left[\frac{x}{3}\right]+\left[\frac{x}{4}\right] \leqslant \frac{x}{2}+\frac{x}{3}+\frac{x}{4} .
\end{array}
$$

Solving this, we get $0 \leqslant x \leqslant 23$.
Therefore, $A_{4} \subseteq\{0,1, \cdots, 23\}$.
When $x=23$,
$$
\left[\frac{x}{2}\right]=11,\left[\frac{x}{3}\right]=7,\left[\frac{x}{4}\right]=5,
$$

satisfying $\left[\frac{x}{2}\right]+\left[\frac{x}{3}\right]+\left[\frac{x}{4}\right]=23$.
Thus, $23 \in A_{4}$.
When $n \geqslant 5$,
$$
\begin{array}{l}
x=\left[\frac{x}{2}\right]+\left[\frac{x}{3}\right]+\cdots+\left[\frac{x}{n}\right], \text { and } x \in \mathbf{Z} . \\
\text { Hence } x=\left[\frac{x}{2}\right]+\left[\frac{x}{3}\right]+\cdots+\left[\frac{x}{n}\right] \\
\leqslant \frac{x}{2}+\frac{x}{3}+\cdots+\frac{x}{n},
\end{array}
$$

which means $x \geqslant 0$.
On the other hand,
$$
\begin{array}{l}
x=\left[\frac{x}{2}\right]+\left[\frac{x}{3}\right]+\cdots+\left[\frac{x}{n}\right] \\
\geqslant\left[\frac{x}{2}\right]+\left[\frac{x}{3}\right]+\left[\frac{x}{4}\right] \\
\geqslant \frac{x-1}{2}+\frac{x-2}{3}+\frac{x-3}{4},
\end{array}
$$

which gives $x \leqslant 23$.
In summary, $A=\cup_{n \geqslant 2} A_{n} \subseteq\{-7,-6, \cdots, 23\}$ is a finite set, with the maximum element being $23 \in A_{4}$ and the minimum element being $-7 \in A_{3}$.