4. Let the set $M=\{1,2, \cdots, 2020\}, A \subseteq M$, and for any element $x$ in set $A$, $4x \notin A$. Then the maximum number of elements in set $A$ is $\qquad$ .
Reference answer: First, construct 404 sets $\{k, 4 k\}: k=1 ; 8,9, \cdots, 31 ; 127,128, \cdots 505$. Second, each number $x$ in set $M$ that is not mentioned in the aforementioned 808 numbers forms a set $\{x\}$, totaling 1212. In total, there are $404+1212=1616$ sets. According to the pigeonhole principle, if set $A$ contains more than 1616 numbers, then there must be two numbers taken from the same set mentioned above, thus there exist $x, 4 x \in A$, which is a contradiction. Therefore, set $A$ contains at most 1616 numbers, and a set that satisfies the condition is $A=\{2,3, \cdots, 7 ; 32,33, \cdots, 126 ; 506,507, \cdots, 2020\}$.