Example 4 Given $n$ $(n \geqslant 2)$ positive integers $x_{1}, x_{2}$, $\cdots, x_{n}$, arrange them in non-decreasing order as $x_{1} \leqslant x_{2} \leqslant \cdots$ $\leqslant x_{n}$. If the sum of these $n$ positive integers equals their product, find the maximum value of $x_{n}$.
Explanation: An integer array that meets the requirements of the problem exists, for example, $2+2=2 \times 2, 1+2+3=1 \times 2 \times 3, 1+1+2+4=1 \times 1 \times 2 \times 4$, etc. From these examples, we can see that the maximum value of $x_{n}$ is exactly $n$.
We estimate the maximum value of $x_{n}$ as a whole.
First, $x_{n-1} \geqslant 2$. Otherwise, if $x_{n-1}=1$, then $x_{1}=x_{2}=\cdots=x_{n-1}=1$. Thus,
$$
\begin{array}{l}
x_{1}+x_{2}+\cdots+x_{n}=(n-1)+x_{n} \\
>x_{n}=x_{1} x_{2} \cdots x_{n} .
\end{array}
$$

Next, we determine the upper bound of $x_{n}$. If $x_{n} \geqslant n+1$, since $x_{n-1} \geqslant 2$, then we have
$$
\begin{aligned}
1= & \frac{x_{1}+x_{2}+\cdots+x_{n}}{x_{1} x_{2} \cdots x_{n}} \\
= & \frac{1}{x_{2} x_{3} \cdots x_{n}}+\frac{1}{x_{1} x_{3} \cdots x_{n}}+\cdots \\
& +\frac{1}{x_{1} x_{2} \cdots x_{n-1}} \\
\leqslant & \frac{1}{2(n+1)}+\frac{1}{2(n+1)}+\cdots+\frac{1}{n+1}+\frac{1}{2} \\
= & \frac{n-2}{2(n-1)}+\frac{1}{n+1}+\frac{1}{2}=\frac{2 n+1}{2 n+2}<1 .
\end{aligned}
$$

A contradiction arises. Therefore, $x_{n} \leqslant n$.
We can construct an instance where $x_{n}=n$:
$$
x_{1}=x_{2}=\cdots=x_{n-2}=1, x_{n-1}=2, x_{n}=n \text {, }
$$

then
$$
\begin{array}{l}
x_{1}+x_{2}+\cdots+x_{n}=(n-2)+2+n=2 n \\
=1 \times 1 \times \cdots \times 1 \times 2 \times n=x_{1} x_{2} \cdots x_{n-1} x_{n} .
\end{array}
$$