Given any odd prime $p$, let $r=\lfloor p/2\rfloor-1$. We work in $\mathbb{Z}_p^\ast$ and consider the elements $\pm 1,\pm 1/2,\pm 1/3,\ldots \pm 1/r$. By the pigeonhole principle there is a residue $\bar q$ which is not on the list. We then use Dirichlet to find a prime $q\in\mathbb{N}$ congruent to $\bar q$  modulo $p$.  Then letting $n=pq$, we see that we can't improve the $\frac{n}{2}$ bound significantly for a product of two primes.

For a product of three primes e.g., $2pq$, we can do a little better because solving the problem for $n=pq$ automatically produces a solution for $2pq$ as well.