Three. (50 points) Given $n \in \mathbf{N}, n>4$, the set $A=$ $\{1,2, \cdots, n\}$. Let the positive integer $m$ and the sets $A_{1}, A_{2}, \cdots$, $A_{m}$ have the following properties:
(1) $\bigcup_{i=1}^{m} A_{i}=A$;
(2) $\left|A_{i}\right|=4(i=1,2, \cdots, m)$;
(3) $X_{1}, X_{2}, \cdots, X_{\mathrm{C}_{n}^{2}}$ are all the two-element subsets of $A$, for any $X_{k}\left(k=1,2, \cdots, \mathrm{C}_{n}^{2}\right)$, there exists a unique $j_{k} \in\{1,2, \cdots, m\}$, such that $X_{k} \subseteq A_{j_{k}}$.
Find the minimum value of $n$.
Three, note that the number of binary subsets of set $A$ is $\mathrm{C}_{n}^{2}$; in each $A_{i}$, there are $\mathrm{C}_{4}^{2}$ distinct binary subsets.
From (3) we know
$$
\begin{array}{l}
m \mathrm{C}_{4}^{2}=\mathrm{C}_{n}^{2} \Rightarrow \mathrm{C}_{4}^{2} \mid \mathrm{C}_{n}^{2} \\
\Rightarrow 6 \left\lvert\, \frac{n(n-1)}{2} \Rightarrow 12 \ln (n-1)\right. \\
\Rightarrow n \equiv 0,1,4,9(\bmod 12) .
\end{array}
$$

For any $a \in A$, the number of binary subsets of set $A$ that include $a$ is $n-1$; in each $A_{i}$ that includes $a$, there are 3 distinct binary subsets that include $a$. From (3) we know
$$
3 \mid(n-1) \Rightarrow n \equiv 1(\bmod 3) \text {. }
$$

In summary, $n \equiv 1$ or $4(\bmod 12)$.
Since $n>4$, we have $n \geqslant 13$.
For $A=\{1,2, \cdots, 13\}$, take
$$
A_{i}=\{i, i+1, i+3, i+9\}(i=1,2, \cdots, 13) \text {, }
$$

where the elements are taken modulo 13.
It is easy to verify that $\left|A_{1} \cap A_{i}\right| \leqslant 1(i=2,3, \cdots, m)$.
By translation symmetry, for $1 \leqslant i<j \leqslant m$ we have
$\left|A_{i} \cap A_{j}\right| \leqslant 1$.
Also, $13 \mathrm{C}_{4}^{2}=\mathrm{C}_{13}^{2}$, so each binary subset of set $A$ is contained in exactly one subset $A_{k}(k \in\{1,2, \cdots, m\})$.
Therefore, the minimum value of $n$ is 13.