Four, (50 points) A town has $n(n \geqslant 3)$ villages, some of which are connected by roads, and each road connects only two villages. Any two villages are connected by at most one road, and there are no crossroads at road intersections, but there are bridges or tunnels. It is known that any two villages can be reached via the roads, but after closing any loop due to a bicycle race, the aforementioned conclusion no longer holds. Find the maximum number of roads.
Four, the maximum value is $2n-3$.
First, give the construction: take out two villages that are connected to all other villages, and these two villages are also connected to each other, while the remaining $n-2$ villages are not connected to each other. It is known that this satisfies the requirement and has a total of $2n-3$ roads.
Next, prove that there cannot be more roads.
Consider each village as a vertex and each road as an edge to form an $n$-order graph $G$, i.e., prove that when the number of edges is not less than $2n-2$, it is always possible to remove a cycle so that the remaining graph is still connected.

Clearly, it is only necessary to prove the case with exactly $2n-2$ edges (if the number of edges is greater than $2n-2$, then one can remove one edge from a cycle each time until there are $2n-2$ edges).
When $n=3$, it is obviously true.
Assume that the proposition holds for $n=k$.
Consider the case when $n=k+1$. At this point, there are $2k$ edges. By the pigeonhole principle, there exists a vertex $A$ whose degree does not exceed $\left[\frac{2 \times 2k}{k+1}\right]=3$.
(1) If the degree of vertex $A$ is 1, then remove $A$ and this edge from graph $G$, resulting in a connected graph with $k$ vertices and $2k-1$ edges. By the induction hypothesis, a cycle can be removed while still maintaining connectivity, thus the proposition holds for graph $G$.
(2) If the degree of vertex $A$ is 2, let $A$ be connected to $B$ and $C$. Remove $A$ and $AB$, $AC$ from graph $G$. If $B$ and $C$ were originally connected, leave it unchanged; otherwise, connect $B$ and $C$. At this point, the remaining graph is connected and has at least $2k-2$ edges. Thus, by the induction hypothesis, a cycle can be removed from the remaining graph while maintaining connectivity. If this cycle does not contain points $B$ and $C$, it meets the requirement; if this cycle contains points $B$ and $C$, replace it with $AB$ and $AC$ in the original graph, which also meets the requirement. Therefore, the proposition holds for graph $G$.
(3) If the degree of vertex $A$ is 3, let $A$ be connected to $B$, $C$, and $D$. If $B$, $C$, and $D$ are connected by three edges, then removing these three edges still keeps the graph connected, and the proposition holds.

If $B$, $C$, and $D$ are connected by two edges, then add the third edge. If the cycle does not contain points $B$ and $C$, it meets the requirement; if the cycle contains points $B$ and $C$, replace it with $AB$ and $AC$ in the original graph, which also meets the requirement. Remove $A$ and $AB$, $AC$, $AD$. Similarly, the proposition holds.

If $B$, $C$, and $D$ are connected by one edge or not at all, if $B$ and $C$ are connected, $C$ and $D$ are connected, then in the new graph, replace $BC$ with $AB$ and $AC$, $BD$ with $AB$ and $AD$. If both $BC$ and $BD$ are taken, replace them with $AC$ and $AD$. Then increase the number of edges between them to two and remove $A$ and $AB$, $AC$, $AD$. Similarly, the proposition holds.
In summary, the proposition is proven.