Example 7 Let $S=\{1,2, \cdots, 15\}$, and take $n$ subsets $A_{1}, A_{2}, \cdots, A_{n}$ from $S$, satisfying the following conditions:
(i) $\left|A_{i}\right|=7, i=1,2, \cdots, n$;
(ii) $\left|A_{i} \cap A_{j}\right| \leqslant 3,1 \leqslant i<j \leqslant n$;
(iii) 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 $n$.
Let $A=\left\{A_{1}, A_{2}, \cdots, A_{n}\right\}$ be any family of sets that meets the conditions of the problem. For any $a \in S$, we denote the number of sets in the family $A$ that contain $a$ by $d(a)$. These $d(a)$ sets each contain $C_{6}^{2}=15$ three-element subsets that include $a$. On the other hand, the total number of three-element subsets of $S$ that contain $a$ is $C_{11}^{2}=91$. By (iii), each of these subsets is contained in some set $A_{i}$ in the family $A$, so $15 \cdot d(a) \geqslant 91, d(a) \geqslant 7$.

Now, we calculate the total number of elements in all sets of the family $A$ in two ways, and we get
$$
n \times 7=\sum_{k=1}^{n}\left|A_{k}\right|=\sum_{a \in S} d(a) \geqslant 15 \times 7, n \geqslant 15 .
$$

Next, we construct a specific family of sets $A=\left\{A_{1}, A_{2}, \cdots, A_{15}\right\}$ that meets the conditions of the problem.
First, we mark the 15 elements of $S=\{1,2, \cdots, 15\}$ in a clockwise direction on 15 equally spaced points on a circle (as shown in Figure (a)).

We define $A_{1}=\{1,2,4,5,6,11,13\}$, and denote the 7-element subset of $S$ obtained by rotating $A_{1}$ clockwise by $j-1$ arc segments as $A_{j}(j=2,3, \cdots, 15)$. Here, an arc segment is the arc between two adjacent points on the equally divided circle. Figure (b) illustrates $A_{1}$. The numbers marked on the inner side of the arcs in the figure represent the number of arc segments in that arc.
We will verify that such an $A=\left\{A_{1}, A_{2}, \cdots, A_{15}\right\}$ meets the problem's requirements.
(i) Clearly, each set in $A$ consists of exactly 7 elements of $S$.
(ii) If $1 \leqslant j-i \leqslant 7$, then rotating $A_{i}$ clockwise by $j-i$ arc segments results in $A_{j}$.
Analyzing the diagram of $A_{1}$ (Figure (b)), we see that there are exactly three pairs of elements that are separated by 1, 2, 3, 4, 5, 6, and 7 arc segments, respectively. Each $A_{i}$ has the same property. Rotating $A_{i}$ clockwise by $j-i$ arc segments $(1 \leqslant j-i \leqslant 7)$, exactly three elements of $A_{i}$ will move to positions marked by elements that are still in $A_{i}$, so $\left|A_{i} \cap A_{j}\right|=3$.

If $8 \leqslant j-i \leqslant 14$, then rotating $A_{j}$ clockwise by $15-(j-i)$ arc segments results in $A_{i}$. Similarly, we can conclude that $\left|A_{i} \cap A_{j}\right|=3$.
(iii) Let $M=\{u, v, w\}$ be any three-element subset of $S'$. Without loss of generality, assume $u, v, w$ are arranged in a clockwise direction on the circle. Consider the number of arc segments between $u$ and $v$, $v$ and $w$, and $v$ and $u$ in the clockwise direction, and denote the pair of the smaller two numbers of these arc segments as $(a, b)$ (in clockwise order). We will classify the three-element subsets of $S$ based on these pairs. 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 $A_{1}$, we see that each type of $(a, b)$ pair appears exactly once in $A_{1}$:
$$
\begin{array}{l}
(1,1),(1,2),(2,1),(1,3),(3,1),(1,4),(4,1),(1,5), \\
(5,1),(1,6),(6,1),(1,7),(7,1) ; \\
(2,2),(2,3),(3,2),(2,4),(4,2),(2,5),(5,2),(2,6), \\
(6,2),(2,7),(7,2) ; \\
(3,3),(3,4),(4,3),(3,5),(5,3),(3,6),(6,3),(3,7), \\
(7,3),(4,4),(4,5),(5,4),(5,5) .
\end{array}
$$

Accordingly, any three-element subset $M$ of $S$ is contained in at least one set in the family $A$ (in fact, exactly one set in the family $A$).
In summary, the minimum value of $n$ is 15.