Example 1 Given that the sum of several positive integers is 1976. Find the maximum value of their product.
Explanation: This is a problem from the 18th International Mathematical Olympiad held in 1976, but the year 1976 is not crucial. Replacing 1976 with 2002, 2003, the method remains the same.

Since there are only a finite number of different sets of positive integers whose sum is 1976, and their products are also finite, the maximum product exists.

Let \( n \) positive integers \( x_{1}, x_{2}, \cdots, x_{n} \) sum to 1976, i.e.,
\[
x_{1} + x_{2} + \cdots + x_{n} = 1976.
\]

Here, \( n \) is a variable because the number of positive integers whose sum is 1976 is not fixed. Our goal is to maximize the product, regardless of the number of positive integers \( n \).

First, focus on a positive integer greater than 4.
If \( x_{1}, x_{2}, \cdots, x_{n} \) contains a number greater than 4, say \( x_{j} > 4 \), decompose \( x_{j} \) into the sum of 2 and \( x_{j} - 2 \):
\[
x_{j} = 2 + (x_{j} - 2).
\]

The product of the two addends is
\[
2(x_{j} - 2) = 2x_{j} - 4 = x_{j} + (x_{j} - 4) > x_{j}.
\]

Therefore, the first adjustment is to replace all numbers \( x_{j} \) in \( x_{1}, x_{2}, \cdots, x_{n} \) that are greater than 4 by decomposing them into 2 and \( x_{j} - 2 \), ensuring all numbers are no greater than 4.

Of course, avoid decomposing to 1, as this does not increase the product. Also, note that if the decomposition results in 4, since \( 4 = 2 + 2 = 2 \times 2 \), replacing 4 with \( 2 + 2 \) does not decrease the product.

Thus, 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 into several 2s and 3s.
Next, focus on the adjustments between 2 and 3.
Since \( 2 + 2 + 2 = 3 \times 2 \), but \( 2 \times 2 \times 2 < 3 \times 3 \), this indicates that having more 3s in the decomposition of 1976 is better than having more 2s.

Therefore, the third adjustment is to replace every three 2s in the decomposition of 1976 with two 3s, i.e., to increase the number of 3s.

Since \( 1976 = 658 \times 3 + 2 \), after these three adjustments, 1976 is decomposed into 658 threes and one 2.
At this point, the product is maximized, and the maximum value is \( 2 \times 3^{658} \).

The solution process for this problem is:
\[
\rightarrow \text{A set of positive integers sum to } 1976 \xrightarrow[\text{Decompose numbers greater than 4 into 2, 3, 4}]{\text{First adjustment}}
\]
\[
\text{Several 2s, 3s, 4s sum to } 1976 \frac{\text{Second adjustment}}{4 \text{ decomposed into } 2 + 2}
\]
\[
\text{Several 2s, 3s sum to } 1976 \underset{3 \text{ 2s decomposed into } 2 \uparrow 3}{\text{Third adjustment}}
\]
\[
658 \text{ threes and 1 two sum to } 1976
\]
\[
\text{Maximum product } 2 \times 3^{658}
\]