Three. (50 points) Given that $n$ is a positive integer greater than 10, and set $A$ contains $2n$ elements. If the family of sets
$$
\left\{A_{i} \subseteq A \mid i=1,2, \cdots, m\right\}
$$

satisfies the following two conditions, it is called "suitable":
(1) For any $i=1,2, \cdots, m, \operatorname{Card}\left(A_{i}\right)=n$;
(2) For any $1 \leqslant i < j \leqslant m, \operatorname{Card}\left(A_{i} \cap A_{j}\right) \leqslant 10$,

Given $n>10$, find the maximum positive integer $m$, such that there exists a suitable family of sets containing $m$ sets.
The maximum positive integer $m=4$.
Take two non-complementary $n$-element subsets $A_{1}, A_{2}$ of $A$, and let $A_{3}, A_{4}$ be the complements of $A_{1}, A_{2}$, respectively. From these four sets, any three sets will always have two sets that are complementary, hence the intersection of these three sets is the empty set.

Assume the family of sets $\left\{A_{i} \subseteq A\right\}$ contains 5 subsets of $A$. The 5 subsets $A_{i}(i=1,2, \cdots, 5)$ contain a total of $5 n$ elements.
(1) If there is 1 element that appears 5 times, then the other elements appear at most 2 times. Thus, the 5 subsets contain at most $5 + 2(2 n-1) = 4 n + 3$ elements, which is a contradiction.
(2) If there are 2 elements each appearing 4 times, then these 2 elements must appear together in at least 3 subsets, which is a contradiction.
(3) If exactly 1 element appears 4 times, and let the number of elements appearing 3 times be $s$, then
$$
\begin{array}{l}
4 + 3 s + 2(2 n - s - 1) \geq 5 n \\
\Rightarrow s \geq n - 2 .
\end{array}
$$
(i) When $s > n - 2$, at least $n$ elements each appear at least 3 times;
(ii) When $s = n - 2$, there is 1 element appearing 4 times, and $n - 2$ elements each appearing 3 times.
(4) If no element appears 4 times, and let the number of elements appearing 3 times be $s$. Then
$$
3 s + 2(2 n - s) \geq 5 n \Rightarrow s \geq n .
$$

In summary, there are only two cases:
(1) At least $n$ elements each appear at least 3 times;
(2) One element appears 4 times, and $n - 2$ elements each appear 3 times.

For (1), since the intersection of any 3 subsets out of the 5 subsets contains at most 1 element, the number of elements appearing at least 3 times, $n \leq \mathrm{C}_{5}^{3} = 10$, which is a contradiction.

For (2), similarly, by the number of elements appearing at least 3 times, $n - 1 \leq \mathrm{C}_{5}^{3} = 10$, we know $n = 11$ and the intersection of any 3 subsets out of the 5 subsets is exactly a singleton set. These 10 singleton sets are all different, but for the element appearing 4 times, it appears in 4 singleton sets, which is a contradiction.
Therefore, there does not exist a suitable family of sets containing 5 subsets.
In conclusion, the maximum positive integer $m=4$.