Suppose we are given $A \subseteq \mathbb{N}$ with $\lim\sup_{n\to\infty}\frac{A\cap\{1,\ldots,n\}}{n} > 0$. For $k\in \mathbb{N}, k\geq 2$ we set $$M_A(k) = \{a\in A: ka \in A\}.$$ Does there exist $k\in \mathbb{N}, k\geq 2$ such that $M_A(k)$ is infinite?
Not necessarily: you can in fact have $M_k(A)=\varnothing$ for all integer $k\ge 2$. This was shown by Besicovitch ("On the density of certain sequences of integers", Math. Ann. 110 (1935), no. 1, 336–341) who has constructed a set (of positive integers) of positive upper density such that none of the elements of the set is divisible by any other element.

$\begingroup$ That's an amazing paper, thanks for making me aware of it! $\endgroup$ – Dominic van der Zypen Jul 22 '17 at 13:57
Besicovitch's result mentioned by Seva is quite hard, while in your question you may simply take $A=\cup [n!+1,2n!]$.

$\begingroup$ Lacunary sets like this were the first thing I thought of  but for your construction, $A=[2,2]\cup[3,4]\cup[7,12]\cup[25,48]\cup\dotsb$, and there are lots of elements of $A$ divisible by each other? $\endgroup$ – Seva Jul 22 '17 at 9:51

6$\begingroup$ But for each fixed $k$, $M_A(k)=A\cap \frac1kA$ is finite $\endgroup$ – Fedor Petrov Jul 22 '17 at 10:18