10. (20 points) Given the set $S=\{1,2, \cdots, 13\}$, the sets $A_{1}, A_{2}, \cdots, A_{k}$ are subsets of $S$, satisfying:
(1) $\left|A_{i}\right|=6(i=1,2, \cdots, k)$;
(2) $\left|A_{i} \cap A_{j}\right| \leqslant 2(1 \leqslant i<j \leqslant k)$.
Find the maximum value of $k$.
10. The maximum value of $k$ is 4.

Construct subsets $A_{1}, A_{2}, A_{3}, A_{4}$ that satisfy the condition $S$:
$$
\begin{array}{l}
A_{1}=\{1,2, \cdots, 6\}, \\
A_{2}=\{1,2,7,8,9,10\}, \\
A_{3}=\{1,3,7,11,12,13\}, \\
A_{4}=\{4,5,8,9,11,12\} .
\end{array}
$$

Below is the proof that $k \leqslant 4$.
On one hand, if there exists an element $a$ that appears in at least three subsets, without loss of generality, let $1 \in A_{1}, 1 \in A_{2}, 1 \in A_{3}$, then
$$
\begin{aligned}
\mid & \left|A_{1} \cup A_{2} \cup A_{3}\right| \\
= & \left|A_{1}\right|+\left|A_{2}\right|+\left|A_{3}\right|-\left|A_{1} \cap A_{2}\right|- \\
& \left|A_{1} \cap A_{3}\right|-\left|A_{2} \cap A_{3}\right|+\left|A_{1} \cap A_{2} \cap A_{3}\right| \\
\geqslant & \left|A_{1}\right|+\left|A_{2}\right|+\left|A_{3}\right|-2-2-2+1 \\
= & \left|A_{1}\right|+\left|A_{2}\right|+\left|A_{3}\right|-5=13 .
\end{aligned}
$$

Thus, $A_{1} \cup A_{2} \cup A_{3}=S$ and the inequality holds with equality if and only if
$$
\begin{array}{l}
\left|A_{1} \cap A_{2}\right|=2,\left|A_{1} \cap A_{3}\right|=2,\left|A_{2} \cap A_{3}\right|=2, \\
A_{1} \cap A_{2} \cap A_{3}=\{1\} .
\end{array}
$$

If $k \geqslant 5$, consider $A_{4}$ and $A_{5}$.
Note that, $A_{4} \subseteq A_{1} \cup A_{2} \cup A_{3},\left|A_{4}\right|=6$, and
$$
\left|A_{4} \cap A_{j}\right| \leqslant 2(j=1,2,3) \text {. }
$$

Then $\left|A_{4} \cap A_{j}\right|=2(j=1,2,3)$, and the pairwise intersections are empty sets.
Otherwise, without loss of generality, let
$$
\{1,2\}=A_{1} \cap A_{2},\{1,3\}=A_{2} \cap A_{3},\{1,4\}=A_{3} \cap A_{1} .
$$

Thus, $A_{4} \subseteq\{5,6, \cdots, 13\}$.
Similarly, $A_{5} \subseteq\{5,6, \cdots, 13\}$.
$$
\begin{array}{l}
\text { Also, }\left|A_{4} \cap A_{5}\right|=\left|A_{4}\right|+\left|A_{5}\right|-\left|A_{4} \cup A_{5}\right| \\
\geqslant\left|A_{4}\right|+\left|A_{5}\right|-9=3,
\end{array}
$$

which contradicts $\left|A_{i} \cap A_{j}\right| \leqslant 2$.
Therefore, $k \leqslant 4$.
On the other hand, if no element $a$ appears in at least three subsets, then
$$
k \leqslant \frac{13 \times 2}{6}=\frac{13}{3},
$$

which means $k \leqslant 4$.
In conclusion, the maximum value of $k$ is 4.