Suppose there exists a town $A$ with $k-1$ one way roads all of which are pointing in one and the same direction from $A$. Town $A$ with all $k-1$ end points of one way roads form a group of $k$ towns that violates the condition of the problem. We conclude that for any town there exist at most $k-2$ one way roads in the town and at most $k-2$ roads out of the town. Therefore the one way roads are at most $n(k-2)$.

We show now that when $n \le 2k-3$, it is possible to have all roads to be one way roads and when $n > 2k-3$ then the maximal number of one way roads equals $n(k-2)$.

Let $n = 2k-3$ and consider the towns to be vertexes of a regular $n$-gon. For every town $A$ let $k-2$ one way roads out of $A$ point to the next $k-2$ towns clockwise. Since $n = 2k-3$ such allocation is possible and every two towns are connected with one way road. Consider a group of $k$ towns and order them clockwise. Between any two neighbors $A$ and $B$ there exists one way road from $A$ to $B$. Therefore the desired round trip exists.

When $n < 2k-3$, it suffices to choose arbitrary $n$ towns from the above construction.

When $n > 2k-3$ we proceed by similar manner. Connect every town $A$ with one way road from $A$ with the next $k-2$ towns. Let all remaining roads be two way roads.

Take arbitrary $k$ towns. Consider a group of $k$ towns and order them clockwise. Between any two neighbors $A$ and $B$ there exists one way road from $A$ to $B$. Therefore the desired round trip exists.

Answer. When $n \le 2k-3$ all roads could be one way and when $n > 2k-3$ there exist at most $n(k-2)$ one way roads.