Example 3 Let $n \in \mathbf{N}_{+}, n \geqslant 2, S$ be an $n$-element set. Find the smallest positive integer $k$, such that there exist subsets $A_{1}, A_{2}, \cdots, A_{k}$ of $S$ with the following property: for any two distinct elements $a, b$ in $S$, there exists $j \in\{1,2, \cdots, k\}$, such that $A_{j} \cap\{a, b\}$ is a one-element subset of $S$.
Solution: Let $S=\{1, 2, \cdots, k\}$. Construct the table
$1:$
If $i \in A_{j}$, then mark the cell at the intersection of the row for $A_{j}$ and the column for $i$ with 1, and mark the rest of the cells with 0.
Consider the sequence of columns in Table 1, $P_{1}, P_{2}, \cdots, P_{n}$.
We will prove: The subsets $A_{1}, A_{2}, \cdots, A_{k}$ of $S$ have the property in the problem if and only if $P_{1}, P_{2}, \cdots, P_{n}$ are all distinct.
Sufficiency.
If $P_{1}, P_{2}, \cdots, P_{n}$ are all distinct, then for any $a, b \in S, a \neq b$, we have $P_{a} \neq P_{b}$. Therefore, in some row (let's say the $j$-th row), the cells in the $a$-th and $b$-th columns contain one 1 and one 0. This indicates that $A_{j} \cap \{a, b\}$ is a singleton set, so $A_{1}, A_{2}, \cdots, A_{k}$ have the property in the problem.
Necessity.
Since for any $a, b \in S, a \neq b$, there exists $j \in \{1,2, \cdots, k\}$ such that $A_{j} \cap \{a, b\}$ is a singleton set, then $P_{a}$ and $P_{b}$ have one 1 and one 0 in the $j$-th row, so $P_{a} \neq P_{b}$. Therefore, $P_{1}, P_{2}, \cdots, P_{n}$ are all distinct.

Notice that there are exactly $2^{k}$ distinct sequences of length $k$ consisting of 0s and 1s. Thus, by the pigeonhole principle and the conclusion we have proven, we know that $2^{k} \geqslant n$.
Therefore, the smallest positive integer is
$$
k=\left\{\begin{array}{ll}
\log _{2} n, & \text { when } \log _{2} n \in \mathbf{N}_{+} ; \\
{\left[\log _{2} n\right]+1,} & \text { when } \log _{2} n \notin \mathbf{N}_{+}
\end{array}\right. \text {. }
$$