A Maryland Problem (Edited)

Assuming Annapolis be the town with the most paths (including paths starting from and paths leading to it). We divide the remaining towns into three group:

• Group $I$ represents the towns whose paths start from Annapolis

• Group $II$ represents the towns whose paths lead to Annapolis

• Group $III$ represents the towns whose paths don't start from or leads to Annapolis

Let $m=|I|,n=|II|,p=|III|$ . We have: $m+n+p=155$

We can imply there is no path within the towns of Group $I$ , as well as Group $II$

Because Annapolis is the town with the most paths (including paths starting from and paths leading to it) so $m+n>m+p$ and $m+n>n+p$

$=>$ the number of possible paths maximise when the towns in Group $III$ connect with the towns in Group $I$ and $II$

The number of paths related with towns in Group $III$ doesn't exceed $p.(m+n)$ (Because we have assumed that Annapolis is the town with the most paths including paths starting from and paths leading to it $(m+n)$ )

The number of possible paths include:

• Paths starting from or leading to Annapolis : $m+n$

• Paths related with towns in Group $III$ $\leq p.(m+n)$

• Paths between Group $I$ and $II$ $\leq m.n$

Therefore $m+n + p.(m+n) +m.n = m.n + m.(p+1) +n.(p+1) \leq \frac{(m+n+p+1)^2}{3} = \frac{(156)^2}{3} = \boxed{8112}$ a.k.a maximum number of possible paths. This happens when there are 52 towns in each group and towns in Group $I$ have paths to towns in Group $II$ , towns in Group $II$ have paths to towns in Group $III$ and towns in Group $III$ have paths to towns in Group $I$