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 $j \notin A_{i}(i \neq j$, and $i, j=1,2, \cdots, n)$.
Try to answer the following questions:
(1) Find $\sum_{i=1}^{n} \operatorname{Card}\left(A_{i}\right)$;
(2) Find the minimum value of $n$.


Three, Solution: (1) Let $A_{1}, A_{2}, \cdots, A_{n}$ be $n$ sets where $r_{i}$ sets contain the element $i (i=1,2, \cdots, n)$. Then $\sum_{i=1}^{n} \operatorname{Card}\left(A_{i}\right)=\sum_{i=1}^{n} r_{i}$. Without loss of generality, assume the $r_{1}$ sets containing element 1 are $A_{2}, A_{3}, \cdots, A_{r_{1}+1}$. By condition (ii), we know $2 \notin A_{1}, 3 \notin A_{1}, \cdots, r_{1}+1 \notin A_{1}, r_{1}+2 \in A_{1}, r_{1}+3 \in A_{1}, \cdots, n \in A_{1}$. Also, by condition (i), we know $1 \notin A_{1}$, so $\operatorname{Card}\left(A_{1}\right)=n-r_{1}-1$. Similarly, $\operatorname{Card}\left(A_{i}\right)=n-r_{i}-1$. Therefore, $\sum_{i=1}^{n} \operatorname{Card}\left(A_{i}\right)=\sum_{i=1}^{n}\left(n-r_{i}-1\right)=\sum_{i=1}^{n}(n-1)-\sum_{i=1}^{n} r_{i}$, so $\sum_{i=1}^{n} r_{i}=\sum_{i=1}^{n}(n-1)-\sum_{i=1}^{n} r_{i}$, thus $\sum_{i=1}^{n} r_{i}=\frac{1}{2} \sum_{i=1}^{n}(n-1)=\frac{n(n-1)}{4}$.
(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 $A_{i} \subseteq\{1,2, \cdots, n\}(i=1,2, \cdots, n)$, so among the $n$ elements $1,2, \cdots, n$, at least one element is in 3 sets. Assume it is 1, i.e., $r_{1} \geqslant 3$. Also, $\operatorname{Card}\left(A_{1}\right)=n-r_{1}-1 \geqslant 3$, so $n \geqslant r_{1}+4 \geqslant 7$.

And $A_{1}=\{2,3,4\}, A_{2}=\{3,4,5\}, A_{3}=\{4,5,6\}, A_{4}=\{5,6,7\}, A_{5}=\{6,7,1\}, A_{6}=\{7,1,2\}, A_{7}=\{1,2,3\}$. Therefore, the minimum value of $n$ is 7.