Solution:

The problem is equivalent, in general, to finding the least number of edges required so that a graph on $n$ vertices will be connected, i.e., one can reach any vertex from any other vertex by following the edges of the graph. (We are letting settlements be vertices and tunnels be edges, of course.) This value is $\binom{n-1}{2} + 1$.

Here $\binom{m}{2}$ counts the number of all possible pairs in a group of $m$ people, or equivalently, the number of edges in a graph with $m$ vertices where every two vertices are connected with an edge. This latter graph is called a complete graph on $m$ vertices.

To see that the minimum number of edges must be $\binom{n-1}{2} + 1$, we first observe that it cannot be less than this, since $n-1$ vertices can be connected to one another with $\binom{n-1}{2}$ edges, leaving the $n$th vertex isolated.

Next we will show that $\binom{n-1}{2} + 1$ edges will guarantee that the graph is connected.

Method 1: Suppose to the contrary, that the graph is not connected. Then it consists of $k$ connected components, each containing $v_{1}, v_{2}, \ldots, v_{k}$ vertices. Each component has at most $\binom{v_{i}}{2}$ edges. We claim that
$$\binom{v_{1}}{2} + \binom{v_{2}}{2} + \cdots + \binom{v_{k}}{2} \leq \binom{n-1}{2}$$
which establishes the contradiction.

Method 2: Since there are at most $\binom{n}{2}$ tunnels possible, there will be at most
$$\binom{n}{2} - \left(\binom{n-1}{2} + 1\right) = n-2$$
tunnels that are not drawn. Call these "antitunnels." Suppose to the contrary, that the graph is not connected. Then two settlements, $A$ and $B$ will not be connected. Thus, $A$ and $B$ are joined by an antitunnel. Furthermore, for each settlement $X$ that is neither $A$ nor $B$, there can be no path drawn from $A$ to $X$ and then from $X$ to $B$. In other words, at least one of the connections $A X$ or $X B$ must be an antitunnel. However, this would require $n-2$ antitunnels, in addition to the antitunnel joining $A$ and $B$. Thus $n-1$ antitunnels are needed, but at most $n-2$ are available; a contradiction.