Three. (50 points) Given sets $A_{1}, A_{2}, \cdots, A_{n}$ are different subsets of the set $\{1,2, \cdots, n\}$, satisfying the following conditions:
(i) $i \notin A_{i}$ and $\operatorname{Card}\left(A_{i}\right) \geqslant 3, i=1,2, \cdots, n$;
(ii) $i \in A_{j}$ if and only if $j \notin A_{i}(i \neq j$, $i, j=1,2, \cdots, n)$.
Try to answer the following questions:
(1) Find $\sum_{i=1}^{n} \operatorname{Card}\left(A_{i}\right)$;
(2) Find the minimum value of $n$.


(1) Let $A_{1}, A_{2}, \cdots, A_{n}$ be $n$ sets, and let $r_{i}$ be the number of sets that contain element $i$ for $i=1,2, \cdots, n$. Then,
$$
\sum_{i=1}^{n} \operatorname{Card}\left(A_{i}\right)=\sum_{i=1}^{n} r_{i} \text {. }
$$

Assume the $r_{1}$ sets that contain element 1 are
$$
A_{2}, A_{3}, \cdots, A_{r_{1}+1} \text {. }
$$

By condition (ii), we know that $2 \notin A_{1}, 3 \notin A_{1}, \cdots \cdots r_{1}+1 \notin A_{1}, r_{1}+2 \in A_{1}, r_{1}+3 \in A_{1}, \cdots \cdots n \in A_{1}$.
Also, by condition (i), we know that $1 \notin A_{1}$.
Thus, $\operatorname{Card}\left(A_{1}\right)=n-\left(r_{1}+2\right)+1=n-r_{1}-1$.
Similarly, $\operatorname{Card}\left(A_{i}\right)=n-r_{i}-1$ for $i=1,2, \cdots, n$.
Therefore, $\sum_{i=1}^{n} \operatorname{Card}\left(A_{i}\right)=\sum_{i=1}^{n}\left(n-r_{i}-1\right)$ $=\sum_{i=1}^{n}(n-1)-\sum_{i=1}^{n} r_{i}=\sum_{i=1}^{n} r_{i}$.
Thus, $\sum_{i=1}^{n} r_{i}=\frac{1}{2} \sum_{i=1}^{n}(n-1)=\frac{1}{2} n(n-1)$.
(2) Since $\operatorname{Card}\left(A_{i}\right) \geqslant 3$, we have $\sum_{i=1}^{n} \operatorname{Card}\left(A_{i}\right) \geqslant 3 n$.
And since $A_{i} \subseteq\{1,2, \cdots, n\}$ for $i=1,2, \cdots, n$, at least one of the elements $1,2, \cdots, n$ must be in at least 3 sets. Assume this element is 1, i.e., $r_{1} \geqslant 3$.
$$
\begin{array}{l}
\text { Also, } \operatorname{Card}\left(A_{1}\right)=n-r_{1}-1 \geqslant 3, \text { so } n \geqslant r_{1}+4 \geqslant 7 \text {. And } \\
A_{1}=\{2,3,4\}, A_{2}=\{3,4,5\}, A_{3}=\{4,5,6\}, \\
A_{4}=\{5,6,7\}, A_{5}=\{6,7,1\}, A_{6}=\{7,1,2\}, \\
A_{7}=\{1,2,3\}
\end{array}
$$

satisfy the conditions.
Therefore, the minimum value of $n$ is 7.