Example 6 Given $n$ positive integers $x_{1}, x_{2}, \cdots, x_{n}$ satisfying $x_{1}+x_{2}+\cdots+x_{n}=2008$. Find the maximum value of the product $x_{1} x_{2} \cdots x_{n}$. ${ }^{[3]}$
(2008, National Junior High School Mathematics Competition, Tianjin Preliminary)
Let the maximum value of $x_{1} x_{2} \cdots x_{n}$ be $M$.
From the given equation, we know that each $x_{i}>1$ $(i=1,2, \cdots, 2008)$. If there is an $x_{i} \geqslant 4$, we can split $x_{i}$ into $x_{i}-2$ and 2, and consider their product, we have
$$
\left(x_{i}-2\right) \times 2=x_{i}+\left(x_{i}-4\right) \geqslant x_{i} \text {. }
$$

This means that after splitting any positive integer $x_{i}$ that is greater than or equal to 4 into $x_{i}-2$ and 2, the sum remains unchanged, but the product increases.

Therefore, in $M$, there cannot be any positive integers greater than or equal to 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 will increase the product, so $p \leqslant 2$.
$$
\begin{array}{l}
\text { Also, } 2008=3 \times 668+2+2 \text {, then } \\
M=2^{2} \times 3^{668} .
\end{array}
$$