Example 5 Sets $S_{1}, S_{2}, \cdots, S_{n}$ are pairwise distinct and satisfy the following conditions:
(1) $\left|S_{i} \cup S_{j}\right| \leqslant 2004(1 \leqslant i, j \leqslant n, i, j \in \mathbf{N}_{+})$;
(2) $S_{i} \cup S_{j} \cup S_{k}=\{1,2, \cdots, 2008\}(1 \leqslant i < j < k \leqslant n, i, j, k \in \mathbf{N}_{+})$.
Find the maximum possible value of $n$. ${ }^{[1]}$
$(2009$, Serbian Mathematical Olympiad)
Let the complement of $S_{i}$ in $Y=\{1,2, \cdots, 2008\}$ be $A_{i}(1 \leqslant i \leqslant n)$. Then
$\left|A_{i} \cap A_{j}\right| \geqslant 4, A_{i} \cap A_{j} \cap A_{k}=\varnothing$.
Let $X=\left\{A_{1}, A_{2}, \cdots, A_{n}\right\}$.
Consider the bipartite graph $X+Y$, where a vertex $A_{i} \in X$ is adjacent to $y \in Y$ if and only if $y$ is an element of $A_{i}$.

Since $\left|A_{i} \cap A_{j}\right| \geqslant 4$, for any two vertices in $X$, their common neighbors (referring to points adjacent to both, the same below) are at least four. In other words, for any two vertices in $X$, there are at least four vertices in $Y$ that form an angle with them. Therefore, the number of angles formed by vertices in $Y$ is greater than or equal to $4 \mathrm{C}_{n}^{2}$.

On the other hand, since $A_{i} \cap A_{j} \cap A_{k}=\varnothing$, the degree of each vertex in $Y$ is less than or equal to 2, meaning that at most one angle is formed at each vertex in $Y$. Therefore, the number of angles formed by vertices in $Y$ is less than or equal to $|Y|=2008$.
In summary, $4 \mathrm{C}_{n}^{2} \leqslant 2008(n \leqslant 32)$.
Using the bipartite graph $X+Y$, it is not difficult to construct an example where $n=32$.
Therefore, the largest positive integer sought is 32.