Example 9 Given that the sum of several positive integers is 1976, find the maximum value of their product. (IMO - 17 problem)
The number of different positive integer solutions to the sum 1976 is finite, so this maximum value exists.
Let $x_{1}, x_{2}, \cdots, x_{n}$ be positive integers, $x_{1}+x_{2}+\cdots+x_{n}=1976$ and maximize their product $P=x_{1} x_{2} \cdots x_{n}$.
(i) For all $i$, $x_{i} \leqslant 4$. If not, suppose $x_{j}>4$, then $x_{j}=2+\left(x_{j}-2\right)$, and $2\left(x_{j}-2\right)=2 x_{j}-4>x_{j}$. Therefore, replacing $x_{j}$ with 2 and $x_{j}-2$ would increase the product, which is impossible.
(ii) For all $i$, $x_{i} \geqslant 2$. If there is some $x_{j}=1$, then $x_{i} x_{j}=x_{i}<x_{i}+x_{j}$. Therefore, replacing $x_{i}, x_{j}$ with $x_{i}+x_{j}$ would increase the product $P$, which is a contradiction.
(iii) Since $4=2+2=2 \cdot 2$, $x_{i}=4$ is unnecessary, as it can be replaced by two 2s without changing the sum or the product.
(iv) From the above arguments, $P=2^{r} \cdot 3^{s}$, where $r$ and $s$ are non-negative integers. Since $2+2+2=3+3$ and $2^{3}<3^{2}$, it must be that $r<3$. Given $1976=658 \cdot 3+2$, we get $r=1, s=658$.
Therefore, the maximum value of $P$ is $2 \cdot 3^{658}$.