Have I understood the problem correctly? If so, then... Let the islands have $t_1, ..., t_{13}$ towns respectively, where $t_1+...+t_{13}=25$. For a town on island $i$, it must be joined to $25-t_i$ towns. So the amount of connections given by island $i$ is $t_i(25-t_i)$ (each connection is counted twice here). So we want to minimize the total, that is $ \frac{t_1(25-t_1)+...+t_{13}(25-t_{13})}{2}$ which is the same as minimizing $t_1(25-t_1)+...+t_{13}(25-t_{13})$. Then the problem restates as: If $t_1, ..., t_{13}$ are positive integers such that $t_1+...+t_{13}=25$, find the minimum of $t_1(25-t_1)+...+t_{13}(25-t_{13})$. Now \[t_1(25-t_1)+...+t_{13}(25-t_{13})=-(t_1^2+...+t_{13}^2)+25(t_1+...+t_{13}) \\ -(t_1^2+...+t_{13}^2)+25(t_1+...+t_{13})=-(t_1^2+...+t_{13}^2)+25^2 \] Since $25^2$ is a constant, we want to minimize $-(t_1^2+...+t_{13}^2)$, that is, to maximize $t_1^2+...+t_{13}^2$. Now we use an optimization argument. Suppose $a \ge b$ and consider $a^2+b^2$. Now note that $(a+1)^2+(b-1)^2=a^2+2a+1+b^2-2b+1=(a^2+b^2)+2(a-b+1)>a^2+b^2$. Thus we can't have two numbers greater than $1$ in an optimal case, because otherwise we would apply that transformation and increase the sum. So WLOG $t_1=...=t_{12}=1, t_{13}=13$. Then the least possible number of connections is $\frac{-181+625}{2}=222$.

The most edges are when the partition is quasi-balanced; the least - when the partition is the most unbalanced. By the way, the problem shuns the more economic way of just establishing $\binom {13} {2} = 78$ ferry connections between the islands - as nothing is done for towns on a same island, they must be considered on en equality foot