Nobody or ideas?

We will prove more general result. Let $ G$ be a connected graph with the minimum degree of vertex at least $ d$ and the longest distance between two vertices at least $ D$ (the diameter of $ G$). Then $ |G| \geq \lceil \frac{D+1}{3} \rceil (d-2) + D + 1$. Indeed, consider the two vertices with distance $ D$ and the path connecting them. It's clear that if we pick two vertices of this path which are at least $ 3$ edges apart then they don't have a common neighbour because otherwise we could've shorten the path. For the same reason no two non consecutive vertices of this path are connected. So if we name the vertices in path by $ 1, 2, ... D+1$ then every vertex $ 1, 4, 7, 10, ...$ gives us at least $ d-2$ new vertices which don't belong to the path. After adding the vertices in the path our inequality follows. Suppose that there are two towns which require at least $ 63$ transfers, i.e. $ D \geq 63$. Since $ d \geq 93$ we obtain the bound: $ 1993 \geq \lceil \frac{63+1}{3} \rceil (93-2) + 63 + 1 = 22 \cdot 91 + 64 = 2066$ which clearly can't hold. It seems to me that we can lower $ d$ to $ 90$ instead of $ 93$...

TomciO wrote: For the same reason no two non consecutive vertices of this path are connected. I assume you don't have to use this, right?

I have to use this because otherwise some of that $ d-2$ vertices which are produced by $ 1, 4, 7, ...$ could appear somewhere in the path.

I see. Very nice.