Example 12 Let $S=\{1,2,3, \cdots, 15\}$, and take $n$ subsets $A_{1}, A_{2}, \cdots, A_{n}$ from $S$ satisfying the following conditions:
(1) $\left|A_{i}\right|=7, i=1,2, \cdots, n$;
(2) $\left|A_{i} \cap A_{j}\right| \leqslant 3(1 \leqslant i<j \leqslant n)$;
(3) For any three-element subset $M$ of $S$, there exists some $A_{k}$ such that $M \subset A_{k}$.
Find the minimum value of the number of such subsets $n$.
(1999 China National Team Selection Examination Problem)
Let $A=\left\{A_{1}, A_{2}, \cdots, A_{n}\right\}$ be any family of sets that satisfies the conditions of the problem.
Construct a $15 \times n$ table, where the element in the $i$-th row and $j$-th column is
$$
x_{i j}=\left\{\begin{array}{l}
1\left(\text { if } i \in A_{j}\right) \\
0\left(\text { if } i \notin A_{j}\right)
\end{array}(i=1,2, \cdots, 15 ; j=1,2, \cdots, n)\right. \text {. }
$$

Thus, the sum of the elements in the $i$-th row $d_{i}=\sum_{j=1}^{n} x_{i i}$ represents the number of sets in $A$ that contain $i$, and $\left|A_{j}\right|=\sum_{i=1}^{15} x_{i j}$ represents the number of elements in $A_{j}$. Therefore, by the given condition (1), we have
$$
7 n=\sum_{j=1}^{n}\left|A_{j}\right|=\sum_{j=1}^{n} \sum_{i=1}^{15} x_{i j}=\sum_{i=1}^{15} \sum_{j=1}^{n} x_{i j}=\sum_{i=1}^{15} d_{i} .
$$
Each set in $A$ that contains $i$ has $C_{6}^{2}=15$ three-element subsets containing $i$, so the total number of three-element subsets containing $i$ in all sets of $A$ is $15 d_{i}$. On the other hand, the number of three-element subsets of $S$ containing $i$ is $C_{14}^{2}=91$. By the given condition (3), each of these subsets is contained in some set $A_{j}$ in $A$, so
$15 d_{i} \geqslant 91$, hence $d_{i} \geqslant 7$.

Substituting into (1) gives $7 n \geqslant 15 \times 7$, so $n \geqslant 15$.
Next, we construct a family of sets $A=\left\{A_{1}, A_{2}, \cdots, A_{15}\right\}$ that satisfies the conditions of the problem. First, mark the 15 elements of $S=\{1,2, \cdots, 15\}$ in a clockwise order on 15 equally spaced points on a circle (as shown in Figure 4-14(a)). Let $A_{1}=\{1,2,4,5,6,11,13\}$, and denote the 7-element subset of $S$ obtained by rotating $A_{1}$ clockwise by $j-1$ unit arcs as $A_{j}(j=2,3, \cdots, 15)$, where a unit arc is one-fifteenth of the circumference. Figure 4-14(b) illustrates $A_{1}$, with the numbers on the inner arcs indicating the number of unit arcs they contain.
We now prove that the constructed $A=\left\{A_{1}, A_{2}, \cdots, A_{15}\right\}$ meets the problem's requirements.
(1) Clearly, each set in $A$ consists of exactly 7 elements of $S$.
(2) If $1 \leqslant j-i \leqslant 7$, then rotating $A_{i}$ clockwise by $j-i$ unit arcs results in $A_{j}$. From the diagram of $A_{1}$ (Figure 4-14(b)), we see that there are exactly 3 pairs of elements separated by unit arcs of lengths 1, 2, 3, 4, 5, 6, and 7, and each $A_{i}$ has the same property. Therefore, rotating $A_{i}$ clockwise by $j-i$ unit arcs results in exactly 3 elements of $A_{i}$ landing in positions where the elements are in $A_{j}$, so $\left|A_{i} \cap A_{j}\right|=3$.

If $8 \leqslant j-i \leqslant 14$, then rotating $A_{j}$ clockwise by $15-(j-i)$ unit arcs results in $A_{i}$, and similarly, we can conclude that $\left|A_{i} \cap A_{j}\right|=3$.
(3) Let $M=\{u, v, w\}$ be any three-element subset of $S$. Without loss of generality, assume $u, v, w$ are arranged clockwise on the circle. Consider the number of unit arcs between $u$ and $v$, $v$ and $w$, and $w$ and $u$, and denote the smaller two numbers of these arcs as $(a, b)$ (in clockwise order). Clearly, $1 \leqslant a, b \leqslant 7$, and except for the pair $(5,5)$, all other pairs $(a, b)$ satisfy $a+b \leqslant 9$.

Observing the diagram of $A_{1}$ (Figure 4-14(b)), we see that each type of $(a, b)$ three-element set appears exactly once in $A_{1}$:
$$
(1,1),(1,2),(2,1),(1,3),(3,1),(1,4),(4,1),(1,5),(5,1),(1,6),(6,1),(1,7),(7,1);
(2,2),(2,3),(3,2),(2,4),(4,2),(2,5),(5,2),(2,6),(6,2);
(3,3),(3,4),(4,3),(3,5),(5,3),(3,6),(6,3);
(4,4),(4,5),(5,4),(5,5).
$$
Accordingly, any three-element subset $M$ of $S$ is contained in at least one set in $A$ (for example, the three-element subset $\{3,8,11\}$ of $S$ corresponds to the ordered pair $(5,3)$, and the three-element subset in $A_{1}$ corresponding to the ordered pair $(5,3)$ is $\{11, 1, 4\}$. Since $11-3=8$, we know that $A_{8}=\{8,9,11,12,13,3,5\}$ contains the three-element subset $\{3,8,11\}$. In fact, each three-element subset of $S$ is contained in exactly one set in $A$).
In summary, the minimum value of $n$ is 15.