Four, (50 points) Given five-element sets $A_{1}, A_{2}, \cdots, A_{10}$, any two of these ten sets have an intersection with at least two elements. Let $A=\bigcup_{i=1}^{10} A_{i}=\left\{x_{1}, x_{2}, \cdots, x_{n}\right\}$, for any $x_{i} \in A$, the number of sets among $A_{1}, A_{2}, \cdots, A_{10}$ that contain the element $x_{i}$ is $k_{i}(i=1,2, \cdots, n)$, and let $m=$ $\max \left\{k_{1}, k_{2}, \cdots, k_{n}\right\}$. Find the minimum value of $m$.
Four, it is easy to get $\sum_{i=1}^{n} k_{i}=50$.
The $k_{i}$ sets containing $x_{i}$ form $\mathrm{C}_{k_{i}}^{2}$ set pairs, $\sum_{i=1}^{n} \mathrm{C}_{k_{i}}^{2}$ includes all set pairs, which contain repetitions.

From the fact that the intersection of any two sets among $A_{1}, A_{2}, \cdots, A_{10}$ has at least two elements, we have
$$
\begin{array}{l}
\sum_{i=1}^{n} \mathrm{C}_{k_{i}}^{2} \geqslant 2 \mathrm{C}_{10}^{2}=90 . \\
\text { Hence } 180 \leqslant \sum_{i=1}^{n} k_{i}\left(k_{i}-1\right) \\
\leqslant \sum_{i=1}^{n} k_{i}(m-1)=45(m-1) \\
\Rightarrow m \geqslant 5 .
\end{array}
$$

Below is the construction for $m=5$.
$$
A_{i}=\{\bar{i}, \overline{i+1}, \overline{i+2}, \overline{i+4}, \overline{i+7}\},
$$

where $i=1,2, \cdots, 10, \bar{i} \equiv i(\bmod 10)$, and $\bar{i} \in$ $\{1,2, \cdots, 10\}$