3. Let $S=\{1,2, \cdots, 10\}, A_{1}, A_{2}, \cdots, A_{k}$ be subsets of $S$ that satisfy: (1) $\left|A_{i}\right|=5, i=1,2, \cdots, k_{i}$ (2) $\mid A_{i} \cap$ $A_{j} \mid \leqslant 2, i, j=1,2, \cdots, k, i \neq j$. Find the maximum value of $k$. (6th test, 1st question, provided by Wu Chang)
First, prove that each element in $S$ belongs to at most 3 of the sets $A_{1}, A_{2}, \cdots, A_{k}$.

By contradiction, suppose there is an element in $S$ that belongs to at least 4 subsets. Without loss of generality, assume $1 \in A_{1}, A_{2}, A_{3}, A_{4}$.
(1) If each of the other elements in $S$ belongs to at most 2 of the sets $A_{1}, A_{2}, A_{3}, A_{4}$, each $A_{i} (i=1,2,3,4)$ has 5 positive integers. Removing 1, there are 4 positive integers left. Thus, excluding 1, $A_{1}, A_{2}, A_{3}, A_{4}$ should have $4 \times 4 = 16$ positive integers, some of which are repeated. However, $S - \{1\}$ consists of 9 different positive integers, so there should be $16 - 9 = 7$ repeated positive integers. Given the initial assumption, 7 positive integers belong to exactly 2 of these 4 sets. Therefore, using the problem's condition, we have
$$
2 C_{4}^{2} \geqslant \sum_{1<i<j<4}\left|A_{i} \cap A_{j}\right| \geqslant C_{4}^{2} + 7,
$$

but $2 C_{4}^{2} = 12$ and $C_{4}^{2} + 7 = 13$, which is a contradiction.
(2) If there is another element in $S$ that belongs to at least 3 of the sets $A_{1}, A_{2}, A_{3}, A_{4}$, without loss of generality, assume $2 \in A_{1} \cap A_{2} \cap A_{3}$. Since each of $A_{1}, A_{2}, A_{3}$ has 5 positive integers, removing 1 and 2, each $A_{i} (i=1,2,3)$ has 3 positive integers left. Thus, excluding 1 and 2, $A_{1}, A_{2}, A_{3}$ should have $3 \times 3 = 9$ positive integers, but $S - \{1, 2\}$ consists of 8 different positive integers. Therefore, $9 - 8 = 1$, meaning there is at least one positive integer, which is neither 1 nor 2, belonging to 2 of the sets $A_{1}, A_{2}, A_{3}$. Thus, we have
$$
6 = 2 C_{3}^{2} \geqslant \sum_{1<i<j<3}\left|A_{i} \cap A_{j}\right| \geqslant 2 \times 3 + 1 = 7,
$$

which is a contradiction.
Thus, $k \leqslant \frac{3 \times 10}{5} = 6$.
On the other hand, let
$$
\begin{array}{l}
A_{1}=\{1,2,3,4,5\}, \quad A_{2}=\{1,2,6,7,10\} \\
A_{3}=\{1,3,8,9,10\}, \quad A_{4}=\{2,4,5,8,9\} \\
A_{5}=\{3,5,6,7,8\}, \quad A_{6}=\{4,5,7,9,10\}.
\end{array}
$$