Three. (50 points) Given $n$ four-element sets $A_{1}, A_{2}, \cdots, A_{n}$, any two of which have exactly one common element, and
$$
\operatorname{Card}\left(A_{1} \cup A_{2} \cup \cdots \cup A_{n}\right)=n .
$$

Find the maximum value of $n$. Here $\operatorname{Card} A$ is the number of elements in set $A$.
Consider any element $a \in A_{1} \cup A_{2} \cup \cdots \cup A_{n}$.
If each $A_{i}$ contains $a$, then by the given condition, the other elements in each $A_{i}$ are all different. Hence,
$$
\operatorname{Card}\left(A_{1} \cup A_{2} \cup \cdots \cup A_{n}\right)=3 n+1>n,
$$

which contradicts the given condition.
Therefore, there must be an $A_{i}$ that does not contain $a$.
Without loss of generality, assume $a \notin A_{1}$. If the number of sets containing $a$ is greater than or equal to 5, then by the given condition, $A_{1}$ shares a common element with each of these 5 sets (this element is certainly not $a$), and these 5 elements are all different (if they were the same, then this common element would be a common element of 2 sets containing $a$, which would mean these two sets have 2 common elements, contradicting the given condition), thus, Card $A_{1} \geqslant 5$, which is a contradiction. Therefore, the number of sets containing $a$ is less than or equal to 4.
On the other hand, since
Card $A_{1}+$ Card $A_{2}+\cdots+\operatorname{Card} A_{n}=4 n$, each element belongs to exactly 4 sets.

Without loss of generality, assume the sets containing element $b$ are $A_{1}, A_{2}, A_{3}, A_{4}$. By the above conclusion,
$$
\operatorname{Card}\left(A_{1} \cup A_{2} \cup A_{3} \cup A_{4}\right)=3 \times 4+1=13.
$$

If $n>13$, then there exists an element $c \notin A_{1} \cup A_{2} \cup A_{3} \cup A_{4}$. Let the set containing $c$ be $A_{5}$, then $A_{5}$ is not one of $A_{1}, A_{2}, A_{3}, A_{4}$. Therefore, $A_{5}$ does not contain $b$. And $A_{5}$ shares a common element with each of $A_{1}, A_{2}, A_{3}, A_{4}$ (certainly not $b$), these 4 common elements are all different (for the same reason as above), and none of them is $c$, thus, $\operatorname{Card} A_{5} \geqslant 5$, which is a contradiction.
Therefore, $n \leqslant 13$.
$n \leqslant 13$ is possible. For example, it is not hard to verify that the following 13 sets meet the requirements.
$$
\begin{array}{l}
\{0,1,2,3\},\{0,4,5,6\},\{0,7,8,9\},\{0,10,11,12\}, \\
\{10,1,4,7\},\{10,2,5,8\},\{10,3,6,9\},\{11,1,5,9\}, \\
\{11,2,6,7\},\{11,3,4,8\},\{12,1,6,8\},\{12,2,4,9\}, \\
\{12,3,5,7\}.
\end{array}
$$

Thus, the maximum value of $n$ is 13.
(Wang Cuixi provided)