Problem 1884

Solution:

Each set $S_{i}$ has at most 2003 elements. Indeed, if $\left|S_{i}\right|=2004$, then from condition $1^{\circ}$ it follows that $S_{j} \subset S_{i}$ for all $j$, contrary to condition $2^{\circ}$. Consider the sets
$$G_{\{i, j\}}=\{1,2, \ldots, 2008\} \backslash\left(S_{i} \cup S_{j}\right) \quad \text{ for } 1 \leq i, j \leq n$$
Then $\left|G_{\{i, j\}}\right| \geqslant 4$ and all $\binom{n}{2}$ sets $G_{\{i, j\}}$ are pairwise disjoint (otherwise, if $x \in G_{\{i, j\}} \cap G_{\{k, l\}}$, then $x \notin S_{i} \cup S_{j} \cup S_{k} \cup S_{l}$, which is impossible if at least three of $i, j, k, l$ are distinct). It follows that $4\binom{n}{2} \leqslant 2008$, whence $n \leqslant 32$.

We will construct 32 sets satisfying $1^{\circ}$ and $2^{\circ}$. Let us partition the set $\{1,2, \ldots, 2008\}$ arbitrarily into $\binom{32}{2}=496$ (disjoint) sets $G_{\{i, j\}}$, where $\left|G_{\{i, j\}}\right| \geqslant 4$ for $1 \leqslant i, j \leqslant 32$, and define
$$S_{i}=\{1,2, \ldots, 2008\} \backslash \bigcup_{j \neq i} G_{\{i, j\}} \quad \text{ for } i=1, \ldots, 32$$
Condition $1^{\circ}$ is automatically satisfied. Moreover, every $s \in\{1,2, \ldots, 2008\}$ belongs to at most one of the sets $G_{\{p, q\}}$, which means that there exist at most two sets $S_{i}$ that do not contain it (these are $S_{p}$ and $S_{q}$), so condition $2^{\circ}$ is also satisfied. Therefore, the answer is $n=32$.