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

Determine the maximum and minimum values of $S=\left|A_{1} \cup A_{2} \cup \cdots \cup A_{n}\right|$.
【Note】 $|M|$ denotes the number of elements in set $M$.
Given that there is at least one element in $A_{2}$ that is 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 $A_{1}, A_{2}, \cdots, A_{n}\left(A_{i}=\{x \mid x \in T, x \neq i(i=1,2\right.$, $\cdots, n) \mid$ ) satisfy: for any $k$ subsets $A_{i_{1}}$, $A_{i_{2}}, \cdots, A_{i_{k}}$, we have
$$
\begin{array}{l}
A_{i_{1}} \cap A_{i_{2}} \cap \cdots \cap A_{i_{k}} \\
\quad=\left\{x \mid x \in T, x \neq i_{1}, i_{2}, \cdots, i_{k}\right\} .
\end{array}
$$

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 $i \neq j(1 \leqslant i, j \leqslant n)$, we have
$$
\left|A_{i} \cap A_{j}\right| \geqslant n+1-2=n-1 \text {, }
$$

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

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

Let
$$
A_{i}=\{i, n+1, n+2, \cdots, 2 n-1\}(i=1,2, \cdots, n) .
$$

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