4. Let $A_{i}=\{i, i+1, i+2, \cdots, i+59\}(i=1,2, \cdots, 11), A_{11+j}=\{11+j, 12+j, \cdots, 70,1,2, \cdots, j\}(j=1,2, \cdots, 59)$. Among these 70 sets, there exist $k$ sets such that the intersection of any 7 of these sets is non-empty. Find the maximum value of $k$.

In these 70 sets, there exist $k$ sets, where the intersection of any 7 sets is non-empty. Find the maximum value of $k$.
4. Let $A_{i}=\{i, i+1, i+2, \cdots, i+59\}(i=1,2, \cdots, 70)$, where the elements in the set are understood modulo 70, i.e., if $x>70$, replace $x$ with $x-70$. Clearly, $60 \in A_{1}, A_{2}, \cdots, A_{60}$, so $k_{\text {max }} \geqslant 60$. We conjecture that $k_{\text {max }}=60$. This only requires proving that for any $k$ satisfying the condition, $k \leqslant 60$. We use proof by contradiction. If $k \geqslant 61$, we prove that any $k$ sets chosen will have 7 sets whose intersection is the empty set. To find 7 sets $A_{i_{1}}, A_{i_{2}}, \cdots, A_{i_{7}}$ whose intersection is the empty set, it is sufficient that their complements' union is the universal set, i.e., $\bar{A}_{i_{1}} \cup \bar{A}_{i_{2}} \cup \cdots \cup \bar{A}_{i_{7}}=I$. Note that $\bar{A}_{i_{1}}, \bar{A}_{i_{2}}, \cdots, \bar{A}_{i_{7}}$ are all 10-element sets, and $I$ has 70 elements. To make $\bar{A}_{i_{1}}, \bar{A}_{i_{2}}, \cdots, \bar{A}_{i_{7}}$ contain all elements, they should be uniformly distributed in $A_{1}, A_{2}, \cdots, A_{70}$, i.e., $A_{i_{1}}, A_{i_{2}}, \cdots, A_{i_{7}}$ should be uniformly distributed in $A_{1}, A_{2}, \cdots, A_{70}$. Therefore, the 7 sets we are looking for have the form: $A_{i}, A_{10+i}, \cdots, A_{60+i}$. The current problem is whether these 7 sets are all chosen. Note that the common feature of these 7 sets is that their indices' unit digits are all $i$. Since $k \geqslant 61$ sets are chosen, at most 9 sets remain, and they cannot simultaneously contain all 10 possible unit digits $0,1,2, \cdots, 9$, i.e., there exists $0 \leqslant i \leqslant 9$ such that $i$ is not the unit digit of the indices of the remaining 9 sets. That is, the sets $A_{i}, A_{10+i}, \cdots, A_{60+i}$ are all chosen sets. And $A_{i} \cap A_{10+i} \cap \cdots \cap A_{60+i}=\Phi$, a contradiction.