Example 9 Given that the sum of several positive integers is 1976, find the maximum value of their product.
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 $2 \cdot 3^{658}$.