Example 24 Divide 2006 into the sum of several distinct positive integers, and make the product of these integers as large as possible. Find the product.
Given that there are only a finite number of ways to divide 2006 into the sum of several distinct positive integers, there must exist a way that maximizes the product of these integers.

If 1 is used as a factor, the product is clearly not maximized (it is sufficient to add 1 to another factor). To divide 2006 into the sum of several distinct positive integers, the more factors there are, the larger the product. To maximize the number of factors, we divide 2006 into \(2+3+4+\cdots+n\) until the sum \(\geqslant 2006\).

If the sum is 1 more than 2006, the number of factors will decrease by at least 1. To maximize the product, we should remove the smallest 2 and add 1 to the last number (the largest).

If the sum is \(m (m \neq 1)\) more than 2006, then removing the number equal to \(m\) will maximize the product.
\[
\begin{array}{c}
2+3+4+\cdots+n \geqslant 2006, \\
\frac{n(n+1)}{2}-1 \geqslant 2006, \\
n^{2}+n-4014 \geqslant 0 .
\end{array}
\]

Since \(n\) is the smallest positive integer satisfying the inequality (1), \(n=63\) (because \(62^{2}+62-4014=-1080\)). At this time,
\[
\begin{array}{c}
2+3+4+\cdots+63=2015, \\
2015-2006=9 .
\end{array}
\]

Therefore, dividing 2006 into
\[
(2+3+\cdots+8)+(10+11+\cdots+63)
\]
this form maximizes the product of these numbers, which is
\[
2 \times 3 \times \cdots \times 8 \times 10 \times \cdots \times 63 .
\]