8. Try to divide 2004 into the sum of several distinct positive integers, and make the product of these positive integers the largest possible. Find this maximum value.
8. Since the number of partition methods is finite, there must exist a partition method that maximizes the product. Suppose dividing 2004 into unequal positive integers \(a_{1}, a_{2}, \cdots, a_{k}\) \(\left(a_{1}<a_{2}<\cdots<a_{k}\right)\) such that their product is maximized. Let \(S=\left\{a_{1}, a_{2}, \cdots, a_{k}\right\}\) represent the partition, \(\Sigma S\) represent the sum of the numbers in \(S\), and \(\Pi S\) represent the product of the numbers in \(S\). The smallest number in \(S\) is \(a_{1}\), and the largest number is \(a_{k}\). Then \(S\) has the following properties: (1) When the numbers in \(S\) are arranged in ascending order, at most one positive integer in the interval \(\left[a_{1}, a_{k}\right]\) does not belong to \(S\). In fact, if \(a, b \notin S (a<b)\) but \(a-1 \in S, b+1 \in S\), let \(S^{\prime}=(S \backslash\{a-1, b+1\}) \cup\{a, b\}\), then \(\Sigma S^{\prime}=\Sigma S\), and \(\frac{\Pi S}{\Pi S^{\prime}}=\frac{(a-1)(b+1)}{a b}=\frac{a b-(b-a)-1}{a b} < 1\), \(\Pi S < \Pi S^{\prime}\), which contradicts \(\Pi S\) being the maximum; (2) \(a_{1} \neq 1\). Because if \(a_{1}=1\), then let \(S^{\prime}=\left(S \backslash\left\{1, a_{k}\right\}\right) \cup\left\{a_{k}+1\right\}\), thus \(\Sigma S^{\prime}=\Sigma S\), but \(\frac{\Pi S}{\Pi S^{\prime}}=\frac{a_{k}}{a_{k}+1}<1\), which is a contradiction; (3) \(a_{1}=2\) or 3, (i) If \(a_{1}=4\) and \(5 \in S\), then let \(S^{\prime}=(S \backslash\{5\}) \cup\{2, 3\}\), then \(\Sigma S^{\prime}=\Sigma S\), and \(\frac{\Pi S}{\Pi S^{\prime}}=\frac{5}{2 \times 3}<1\), which is a contradiction; (ii) If \(a_{1}=4, j \notin S (5 \leqslant j \leqslant k-1)\) and \(k \in S (k \geqslant 6)\), then let \(S^{\prime} \subset\{S \backslash\{4, k\}\} \cup\{2, 3, k-1\}\), then \(\Sigma S^{\prime}=\Sigma S\), and \(\frac{\Pi S}{\Pi S^{\prime}}=\frac{4 k}{2 \times 3 \times (k-1)}=\frac{4 k}{4 k+2(k-3)}<1\), which is a contradiction; (iii) If \(a_{1} \geqslant 5\), let \(S^{\prime}=\left(S \backslash\left\{a_{1}\right\}\right) \cup\left\{2, a_{1}-2\right\}\), then \(\Sigma S^{\prime}=\Sigma S\), and \(\frac{\Pi S}{\Pi S^{\prime}}=\frac{a_{1}}{2(a_{1}-2)}=\frac{a_{1}}{a_{1}+(a_{1}-4)}<1\), which is a contradiction. From the above (1), (2), (3), if \(a_{1}=3\), then from \(3+4+\cdots+n-k=2004\) we get \(n(n+1)=4014+2k\), thus \(n=63, k=9\), i.e., \(S=\{3,4,8,10,11, \cdots, 63\}\), if let \(S^{\prime}=(S \backslash\{11\}) \cup\{2,9\}\), then \(\Sigma S^{\prime}=\Sigma S\), and \(\frac{\Pi S}{\Pi S^{\prime}}=\frac{11}{2 \times 9}<1\), which is a contradiction. Therefore, it can only be \(a_{1}=2\), at this time, \(S=\{2,3, \cdots, 10,12,13, \cdots, 63\}\) makes the product \(\Pi S=\frac{63!}{11}\) the maximum.