If you require that the group be Hausdorff then the answer is No. In fact this is true of any countable group. Here is a proof.
Let $G$ be a countable group. Assume that it does not have the discrete topology. This means that there is some element $g\in G$ such that for any open neighborhood $\mathcal{O}_g$ with $g\in \mathcal{O}_g$ there is a point $h\
eq g$ with $h\in\mathcal{O}_g$. So in particular $g$ is a limit point. Since group multiplication is a homeomorphism you get that every element of the group is a limit point.
Now since the group is locally compact it is a Baire space, since it is Hausdorff then sets of the form $\mathcal{K}_g=G\setminus \\{g\\}$ are open (ie points are closed) and by the previous paragraph they are also dense.
Thus by Baire category we get that $\cap_{g\in G}\mathcal{K}_g=\emptyset$ is dense, which is a contradiction.
Of course if you do not require Hausdorff then you can put the trivial topology on any set.