Four. (50 points) Take the subset $A_{i}=\left\{a_{i}, a_{i+1}, \cdots, a_{i+59}\right\}(i=1,2, \cdots$, 70 ) of the set $S=\left\{a_{1}, a_{2}, \cdots, a_{70}\right\}$, where $a_{70+i}=a_{i}$. If there exist $k$ sets among $A_{1}, A_{2}, \cdots, A_{70}$ such that the intersection of any seven of them is non-empty, find the maximum value of $k$.

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The problem is asking to find the maximum number of sets $k$ from the collection $\{A_1, A_2, \ldots, A_{70}\}$ such that the intersection of any seven of these sets is non-empty. Each set $A_i$ is defined as a subset of $S$ containing 60 consecutive elements, with the indices wrapping around modulo 70.
Given:
$$
A_{1} \cap A_{2} \cap \cdots \cap A_{60}=\left\{a_{60}\right\} \text {, }
$$

Furthermore, the intersection of any seven sets from $A_{1}, A_{2}, \cdots, A_{\infty}$ is non-empty, hence $k \geqslant 60$.

We will now prove that if $k>60$, it cannot be guaranteed that the intersection of any seven sets from the $k$ selected sets is non-empty. Therefore, the maximum value of $k$ is 60.

Divide the known sets $A_{1}, A_{2}, \cdots, A_{70}$ into 10 groups (constructing pigeonholes):
$$
B_{i}=\left\{A_{i}, A_{10+i}, \cdots, A_{60+i}\right\}(i=1,2, \cdots, 10) .
$$

Since the union of the complements $\overline{A_{j}}$ of any seven sets $A_{j}$ within each $B_{i}$ satisfies
$$
\begin{array}{l}
\overline{A_{i}} \cup \overline{A_{10+i}} \cup \cdots \cup \overline{A_{60+i}} \\
=\left\{a_{i+60}, a_{i+61}, \cdots, a_{i+69}\right\} \cup \\
\left\{a_{i+70}, a_{i+11}, \cdots, a_{i+79}\right\} \cup \cdots \cup \\
\left\{a_{i+120}, a_{i+121}, \cdots, a_{i+129}\right\} \\
=\left\{a_{1}, a_{2}, \cdots, a_{70}\right\}=S\left(a_{70+i}=a_{i}\right), \\
\end{array}
$$

it follows that the intersection of any seven sets within each $B_{i}$ is the empty set:
$$
A_{i} \cap A_{10+i} \cap \cdots \cap A_{60+i}=\varnothing \text {. }
$$

For $k>60$, we have $k=6 \times 10+r(1 \leqslant r \leqslant 9)$.
By the pigeonhole principle, any selection of $k$ sets from $A_{1}, A_{2}, \cdots, A_{70}$ must include seven sets from the same $B_{i}(1 \leqslant i \leqslant 10)$.

This contradicts the condition that the intersection of any seven sets is non-empty.
Therefore, the maximum value of $k$ is 60.