Example 11 If the sum of several positive integers is 2011, find the maximum value of their product.
Since there are only a finite number of different sets of positive integers whose sum is 2011, the maximum value must exist.
Let $x_{1}, x_{2}, \cdots, x_{n}$ all be positive integers, $x_{1}+x_{2}+\cdots+x_{n}=2011$ and the product $u=$ $x_{1} x_{2} \cdots x_{n}$ reaches its maximum value, then
(1) $x_{i} \leqslant 4(i=1,2, \cdots, n)$. If not, there is some $x_{j}>4$, since $2+\left(x_{j}-2\right)=$ $x_{j}, 2\left(x_{j}-2\right)=x_{j}+\left(x_{j}-4\right)>x_{j}$, we can replace $x_{j}$ with 2 and $x_{j}-2$ to keep the sum unchanged while increasing the product, which is a contradiction!
(2) $x_{i} \geqslant 2(i=1,2, \cdots, n)$. If not, there is some $x_{k}=1$, by $x_{k} \cdot x_{i}=x_{i}3$, then $u=2^{3} \cdot 2^{\alpha-3} \cdot 3^{\beta}<3^{2} \cdot 2^{\alpha-3} \cdot 3^{\beta}$, and $2+$ $2+2=3+3$ does not change the sum, but $u$ becomes larger, which is a contradiction!

In summary, since $2011=3 \times 670+1$, hence $u_{\max }=2^{2} \times 3^{669}$.
In general, if the sum is $m$, then
$$u_{\max }=\left\{\begin{array}{l}
3^{s}, m=3 s \\
2^{2} \cdot 3^{-1}, m=3 s+1 \\
2 \cdot 3^{s}, m=3 s+2
\end{array}\right.$$