Example 3 Given $n$ positive integers $x_{1}, x_{2}, \cdots, x_{n}$ satisfying $x_{1}+x_{2}+\cdots+x_{n}=2020$. Find the maximum value of the product $x_{1} x_{2} \cdots x_{n}$ of these $n$ positive integers.
(Adapted from the 2008 National Junior High School Mathematics Competition Tianjin Preliminary Round)
Let the maximum value of $x_{1} x_{2} \cdots x_{n}$ be $M$.
From the problem, we know that in $M$, each $x_{i}\left(x_{i}>1\right)(i=1,2, \cdots, n)$, if there is $x_{i} \geqslant 4$, we can split $x_{i}$ into $x_{i}-2$ and 2. Considering their product, we have
$$
\left(x_{i}-2\right) \times 2=x_{i}+\left(x_{i}-4\right) \geqslant x_{i} \text {. }
$$

Thus, after splitting all such positive integers $x_{i} \geqslant 4$ into $x_{i}-2$ and 2, the sum remains unchanged, but the product increases.

Therefore, in $M$, there cannot be any positive integers not less than 4.

Hence, $x_{i}=2$ or 3. This indicates that $M$ can be written in the form $2^{p} \times 3^{q}$.

Furthermore, since $2+2+2=3+3$, but $2^{3}<3^{2}$, replacing three 2s with two 3s in the product results in a larger product, so $p \leqslant 2$.

Since $2020=3 \times 672+2+2$, the maximum value of $x_{1} x_{2} \cdots x_{n}$ is $2^{2} \times 3^{672}$.