[Example 3.3.2] Given that the sum of several positive integers is 1976, find the maximum value of their product.
Since there are only a finite number of different sets of positive integers that sum to 1976, there are also only a finite number of their products, so the maximum product exists.
Let $n$ positive integers $x_{1}, x_{2}, \cdots, x_{n}$ sum to 1976, i.e.,
$$
x_{1}+x_{2}+\cdots+x_{n}=1976 \text {. }
$$

Here, $n$ is a variable, because the number of positive integers whose sum is 1976 is not fixed in the problem. Our goal is to maximize the product, without being constrained by the number of positive integers $n$.
First, consider a positive integer greater than 4.
If there is a number greater than 4 in $x_{1}, x_{2}, \cdots, x_{n}$, for example, $x_{j}>4$, decompose $x_{j}$ into the sum of a 2 and a $x_{j}-2$,
$$
x_{j}=2+\left(x_{j}-2\right),
$$

their product is
$$
2\left(x_{j}-2\right)=2 x_{j}-4=x_{j}+\left(x_{j}-4\right)>x_{j} .
$$

This shows that if we encounter a positive integer $x_{j}$ greater than 4, decomposing it into a 2 and a $x_{j}-2$ will increase the product.

So, the first adjustment is to decompose all numbers greater than 4 in $x_{i}(i=1,2, \cdots, n)$ into numbers no greater than 4 by splitting 2 and $x_{j}-2$. 
Of course, we cannot let the decomposed numbers include 1, because this will not increase the product.
We also need to note that if the decomposed numbers happen to be 4, since
$$
4=2+2=2 \times 2,
$$

so, when replacing 4 with $2+2$, the product will not decrease.
Therefore, the second adjustment is to replace all 4s in $x_{i}$ with $2+2$.
After these two adjustments, the product will increase, and 1976 will be decomposed entirely into the sum of several 2s and 3s, i.e.,
$$
\sum x_{j}=1976, x_{j}=2 \text { or } 3 \text {. }
$$

Next, focus on the adjustments for 2 and 3.
Clearly,
$$
\begin{array}{l}
2+2+2=3+3, \\
2 \times 2 \times 2<3 \times 3 .
\end{array}
$$

This shows that in the decomposition of 1976, more 3s are better than more 2s.
Thus, the third adjustment is to replace every three 2s in the decomposition of 1976 with two 3s, and to make more 3s appear in the decomposition.
Since
$$
1976=658 \times 3+2
$$

Therefore, after these three adjustments, 1976 is decomposed into the sum of 658 3s and 1 2.
At this point, the product is maximized, with the maximum value being $2 \times 3^{658}$.
The solution process for this problem is:
\begin{tabular}{|c|c|c|c|c|c|c|c|c|}
\hline \begin{tabular}{l} 
a set of \\
integers
\end{tabular} & \begin{tabular}{l} 
first \\
adjustment
\end{tabular} & \begin{tabular}{l} 
several \\
$2,3,4$
\end{tabular} & \begin{tabular}{l} 
second \\
adjustment
\end{tabular} & \begin{tabular}{l}
\begin{tabular}{c} 
several \\
2,3
\end{tabular}
\end{tabular} & \begin{tabular}{l} 
third \\
adjustment
\end{tabular} & \begin{tabular}{l}
\begin{tabular}{l}
658 \\
3s and 1
\end{tabular}
\end{tabular} & \begin{tabular}{l} 
maximum \\
product
\end{tabular} & \\
\hline \begin{tabular}{l} 
sum equals \\
1976
\end{tabular} & \begin{tabular}{l} 
numbers greater \\
than 4 are \\
decomposed into \\
$2,3,4$
\end{tabular} & \begin{tabular}{l} 
sum equals \\
1976
\end{tabular} & \begin{tabular}{l}
\begin{tabular}{l}
4 is decomposed \\
into $2+2$
\end{tabular}
\end{tabular} & \begin{tabular}{l} 
sum equals \\
1976
\end{tabular} & \begin{tabular}{l}
\begin{tabular}{l}
3 2s are \\
decomposed into \\
2 3s
\end{tabular}
\end{tabular} & \begin{tabular}{l} 
sum equals \\
1976
\end{tabular} & & $2 \cdot 3^{658}$ \\
\hline
\end{tabular}