Four, (50 points) Let $n \in \mathbf{N}$ and $n \geqslant 2$, $n$ distinct sets $A_{1}, A_{2}, \cdots, A_{n}$ satisfy $\left|A_{i}\right|=n(i=1$, $2, \cdots, n)$, and for any $k(2 \leqslant k \leqslant n-1)$ of the sets $A_{1}, A_{2}, \cdots, A_{n}$, we have
$$
\left|A_{i_{1}} \cap A_{i_{2}} \cap \cdots \cap A_{i_{k}}\right| \geqslant n+1-k .
$$

Determine the maximum and minimum values of $S=\left|A_{1} \cup A_{2} \cup \cdots \cup A_{n}\right|$.
Given that $A_{2}$ has at least one element not in $A_{1}$, we have
$$
\begin{array}{l}
S=\left|A_{1} \cup A_{2} \cup \cdots \cup A_{n}\right| \\
\geqslant\left|A_{1} \cup A_{2}\right| \geqslant n+1 .
\end{array}
$$

Also, for the set $T=\{1,2, \cdots, n+1\}$, its $n$ subsets
$$
A_{1}, A_{2}, \cdots, A_{n}\left(A_{i}=\{x \mid x \in T, x \neq i(i=1,2, \cdots, \right.
$$
$n)\}$) satisfy that for any $k$ subsets $A_{i_{1}}, A_{i_{2}}, \cdots, A_{i_{k}}$, we have
$$
\begin{array}{l}
A_{i_{1}} \cap A_{i_{2}} \cap \cdots \cap A_{i_{k}} \\
=\left\{x \mid x \in T, x \neq i_{1}, i_{2}, \cdots, i_{k}\right\} .
\end{array}
$$

Thus, $\left|A_{i_{1}} \cap A_{i_{2}} \cap \cdots \cap A_{i_{k}}\right|=n+1-k$.

Therefore, $S_{\text {min }}=n+1$.
For any $i \neq j(1 \leqslant i, j \leqslant n)$, since
$$
\left|A_{i} \cap A_{j}\right| \geqslant n+1-2=n-1,
$$

we have $\left|A_{i} \cup A_{j}\right|=\left|A_{i}\right|+\left|A_{j}\right|-\left|A_{i} \cap A_{j}\right|$
$$
\leqslant 2 n-n+1=n+1 \text {. }
$$

Thus, $A_{2}$ has at most one element not in $A_{1}$, $A_{3}$ has at most one element not in $A_{1} \cup A_{2}$, $A_{4}$ has at most one element not in $A_{1} \cup A_{2} \cup A_{3}$, ..., $A_{n}$ has at most one element not in $A_{1} \cup A_{2} \cup \cdots \cup A_{n-1}$.
Hence, $S=\left|A_{1} \cup A_{2} \cup \cdots \cup A_{n-1}\right|+1$
$$
\begin{array}{l}
\leqslant\left|A_{1} \cup A_{2} \cup \cdots \cup A_{n-2}\right|+2 \leqslant \cdots \\
\leqslant\left|A_{1}\right|+n-1=2 n-1 .
\end{array}
$$

Let $A_{i}=\{i, n+1, n+2, \cdots, 2 n-1\}$, where $i=1,2, \cdots, n$. Then for any $k(2 \leqslant k \leqslant n-1)$ subsets $A_{i_{1}}, A_{i_{2}}, \cdots, A_{i_{k}}$, we have
$$
\left|A_{i_{1}} \cap A_{i_{2}} \cap \cdots \cap A_{i_{k}}\right|=n-1 \geqslant n+1-k .
$$

In summary, $S_{\text {max }}=2 n-1$.