Another example is as follows. Take all odd numbers from the set $\{1,2,\dots,2^n\}$ (there are $2^{n-1}$ of them), and the positive powers of two.
Note that, $x+y\mid xy$ implies, if $x=dx_1,y=dy_1$ with $(x_1,y_1)=1$, $x_1+y_1\mid d$. In particular, if we guarantee that, for the collection above, $x_1+y_1\nmid d$, for any $x,y$, we are done. For this, suppose $x$ and $y$ are both odd. Then, $d,x_1,y_1$ are all odd, hence $x_1+y_1$ is even, thus $x_1+y_1\nmid d$. If $x$ is odd, and $y$ is a power of $2$, then $(x,y)=d=1$, hence, $x_1+y_1$ never divides $d$. Finally, if $x=2^k$ and $y=2^\ell$ for some $k>\ell$, we have that $x+y=2^\ell(2^{k-\ell}+1)$, which does not divide $2^{k+\ell}$, as $2^{k-\ell}+1>1$ is an odd integer.