Example + Let $n(\geqslant 2)$ be a positive integer, and $S$ be an $n$-element set. Find the smallest positive integer $k$ such that there exists a family of subsets $A_{1}, A_{2}, \cdots, A_{k}$ of $S$ with the following property: for any two distinct elements $a, b$ of $S$, there exists $j \in\{1,2, \cdots, k\}$ such that $A_{j} \cap\{a, b\}$ is a singleton.
Without loss of generality, let $S=\{1,2, \cdots, n\}$, we define a matrix $\left(a_{i j}\right)_{k \times n}$, where $a_{i j}$ is defined as follows:
$$
a_{i j}=\left\{\begin{array}{ll}
1, & \text { if } j \in A_{i}, \\
0, & \text { if } j \notin A_{i} .
\end{array}\right.
$$

where $i=1,2, \cdots, k, j=1,2, \cdots, n$, and denote the sequence of 0,1 numbers in the $j$-th column as $P_{j}$.
By the condition, for any $a, b \in S, a \neq b$, there exists $i \in\{1,2, \cdots, k\}$, such that $A_{i} \cap\{a, b\}$ is a singleton set. Thus, $P_{a}$ and $P_{b}$ differ in the $i$-th position, indicating that $P_{1}, P_{2}, \cdots, P_{n}$ are all distinct;

Conversely, if $P_{1}, P_{2}, \cdots, P_{n}$ are all distinct, then for any $a, b \in S, a \neq b$, we have $P_{a} \neq P_{b}$, thus $P_{a}$ and $P_{b}$ differ in some component (let's say the $i$-th component), indicating that $A_{i} \cap\{a, b\}$ is a singleton set.

Since the number of 0,1 sequences with $k$ components is $2^{k}$, combining the discussions from both aspects, we should have $2^{k} \geqslant n$. Therefore, $k \geqslant \left[\log _{2} n\right]$ (where $[x]$ denotes the smallest integer not less than $x$).
In summary, the minimum value of $k$ is $\left[\log _{2} n\right]$.