Example 6 Given that the sum of several positive integers is 1976, find the maximum value of their product.
Solution: There are only a finite number of different sets of positive integers whose sum is 1976, so the maximum value exists.

Let $x_{1}, x_{2}, \cdots, x_{n}$ be positive integers, and $x_{1}+x_{2}+\cdots+x_{n}=1976$, with their product $p = x_{1} x_{2} \cdots x_{n}$ achieving the maximum value.

First, $x_{i} \leqslant 4 (i=1,2, \cdots, n)$. If $x_{i}>4$, then $x_{i}=\left(x_{i}-2\right)+2$, and $\left(x_{i}-2\right) \cdot 2=2 x_{i}-4=\left(x_{i}-4\right)+x_{i}>x_{i}$, so replacing $x_{i}$ with $2, x_{i}-2$ will increase the product, leading to a contradiction.

Second, $x_{i} \geqslant 2 (i=1,2, \cdots, n)$. If $x_{i}=1$, then $x_{i} x_{j}=x_{j}<1+x_{j}=x_{i}+x_{j}$, so replacing $x_{i}, x_{j}$ with $x_{i}+x_{j}$ will increase the product, leading to a contradiction.

Since $4=2+2=2 \cdot 2$, we do not need to consider $x_{i}=4$, as it can be replaced by two 2s.

Combining the above results, we can assume $p=2^{r} \cdot 3^{s}, r, s \geqslant 0$, then by $2+2+2=3+3$, and $2^{3}<3^{2}$, we know $r<3$. Also, $1976=658 \times 3+2$, so $r=1, s=658$, thus $p=2 \times 3^{658}$ is the desired result.