For $S\subset \mathbb{N}$ define the upper density as $D^{\ast }(S)=\limsup_{n\rightarrow \infty }\frac{\left\vert S\cap \{1,2,\ldots,n\} \right\vert }{% \left\vert n\right\vert }.$
Question: Suppose $D^{\ast }(S)>0$. Is there $n\in\mathbb{N}$ such that $(S-n )\cap S$ has positive upper density?
$S-n=\{s-n| s\in S\}$
Suppose that $(S-n)\cap S$ has upper density 0 for all $n\leq N$. Then the density of the set of integers $x$, such that $|[xN, (x+1)N]\cap S|\geq 2$ is also 0, hence the upper density of $S$ is $\leq \frac{1}{N}$. Taking $N\rightarrow\infty$ gives a positive answer to your question.