Example 1 The sum of several positive integers is 1976, find the maximum value of their product.
First, consider the simple cases where the sum of several numbers is $4$, $5$, $6$, $7$, and $8$. The partitions that maximize the product are: $\square$
$$4=2+2,5=2+3,6=3+3,7=2+2+3,8=2+3+3.$$

From this, we conjecture: To maximize the product, the sum in the partition should only contain 2 and 3, and there should be at most two 2s.
Solution: First, the set of positive integer partitions of 1976 is finite, so there must be a positive integer partition that maximizes the product of the numbers.

Assume the partition that maximizes the product is $\left(x_{1}, x_{2}, \cdots, x_{n}\right)$, where $x_{1}+x_{2}+\cdots+x_{n}=1976$. At this point, the product of the numbers in the partition is $P=x_{1} x_{2} \cdots x_{n}$. We will prove that when $P$ is maximized, all $x_{i}$ can have the following properties:
(1) $x_{i} \leqslant 3$.

If there is some $x_{i} \geqslant 4$, then replace $x_{i}$ with two numbers: 2 and $x_{i}-2$, to get a new partition: $\left(x_{1}, x_{2}, \cdots, x_{i-1}, x_{i}-2,2, x_{i+1}, \cdots, x_{n}\right)$. Note that $2(x_{i}-2)=2x_{i}-4 \geqslant x_{i}$, so the adjusted $P$ value does not decrease.
(2) $x_{i} \neq 1$.

If there is some $x_{i}=1$, then in the partition, choose any $x_{j}$ and replace 1 and $x_{j}$ with one number: $1 + x_{j}$, to get a new partition: $\left(x_{1}, x_{2}, \cdots, x_{j-1}, x_{j+1}, \cdots, x_{n}, x_{j}+1\right)$. Note that $1 \cdot x_{j} < 1 + x_{j}$, so the adjusted $P$ value increases.
(3) The number of $x_{i}$ equal to 2 is no more than 2.
If $x_{i}=x_{j}=x_{k}=2$, then replace $x_{i}$, $x_{j}$, and $x_{k}$ with two numbers: 3 and 3, to get a new partition. Note that $2 \times 2 \times 2 < 3 \times 3$, so the adjusted $P$ value increases.

From this, we know that $x_{i}$ is either 2 or 3, and the number of 2s is no more than 2. Note that $1976=658 \times 3 + 2$, so the maximum value of $P$ is $3^{658} \times 2$.