In a country, there are 2018 cities, some of which are connected by roads. Each city is connected to at least three other cities. It is possible to travel from any arbitrary city to any other arbitrary city via one or more roads. Consider for each pair of cities the shortest route between these two cities. What is the maximum number of roads that could appear in such a shortest route?
The largest number of roads that can appear in a shortest route is 1511. We first give a country for which this number is assumed. Divide the cities into 504 groups: two groups of five cities $\left(A_{0}, B_{0}, C_{0}, D_{0}, E_{0}\right)$ and $\left(A_{503}, B_{503}, C_{503}, D_{503}, E_{503}\right)$ and further groups of four cities $\left(A_{i}, B_{i}, C_{i}, D_{i}\right)$ for $1 \leq i \leq 502$. For all $i$ with $0 \leq i \leq 503$: connect $A_{i}$ to $B_{i}$, connect $A_{i}$ to $C_{i}$, connect $B_{i}$ to $D_{i}$, and connect $C_{i}$ to $D_{i}$. Further, connect $E_{0}$ to $A_{0}, B_{0}$, and $C_{0}$. Connect $E_{503}$ to $B_{503}, C_{503}$, and $D_{503}$. For $1 \leq i \leq 502$: connect $B_{i}$ to $C_{i}$. Finally, for $0 \leq i \leq 502$: connect $D_{i}$ to $A_{i+1}$. Now every city is connected to exactly three other cities.
If we now want to go from $A_{0}$ to $D_{503}$, then this must be via all $A_{i}$ and $D_{i}$, because these are the only connections between the different groups and each group is only connected to the previous and next group. Furthermore, for $0 \leq i \leq 503$ the city $A_{i}$ is not connected to $D_{i}$, so the route within group $i$ must go via $B_{i}$ or $C_{i}$. At least three of the four or five cities in such a group are thus visited on the route. In total, the route thus contains at least $3 \cdot 504=1512$ cities and therefore at least 1511 roads.
We then show that a shortest route between two cities never goes through more than 1511 roads. Consider such a shortest route where the cities $A_{0}, A_{1}$, $\ldots, A_{k}$ are visited in sequence. Each of the cities $A_{i}$ is connected to at least one other city besides its neighbors $A_{i-1}$ and $A_{i+1}$, say to $B_{i}$ (here $B_{0}, B_{1}, \ldots, B_{k}$ do not have to be all different cities). Furthermore, $A_{0}$ and $A_{k}$ are also connected to a third city, say $C_{0} \neq B_{0}$ and $C_{k} \neq B_{k}$. If one of the cities $B_{i}$ or $C_{i}$ is equal to one of the cities $A_{j}$, then we could have found a shorter route by going directly from $A_{i}$ to $A_{j}$ (or vice versa), contradiction. Therefore, the cities $B_{i}$ and $C_{i}$ are all different from all cities $A_{j}$.
If one of the cities $B_{i}$ is connected to four cities $A_{j}$, say $B_{i}$ is connected to $A_{i}, A_{m}, A_{n}$, and $A_{p}$ with $i<m<n<p$, then we could shorten the route by going from $A_{i}$ to $B_{i}$ and then to $A_{p}$. We would then skip at least two cities from the original route (namely $A_{m}$ and $A_{n}$) and visit only $B_{i}$ in addition, so this new route would be shorter. Contradiction. The cities $B_{i}$ with $3 \leq i \leq k-3$ are thus at least $\frac{k-5}{3}$ different cities. For $B_{0}$ and $C_{0}$, we can shorten the route if one of these cities is equal to $B_{i}$ with $i \geq 3$. That would be a contradiction, so $B_{0}$ and $C_{0}$ are two different cities different from $B_{i}$ with $i \geq 3$. Similarly, $B_{k}$ and $C_{k}$ are two different cities different from $B_{i}$ with $i \leq k-3$. In total, we have $k+1$ cities on the route itself and at least $\frac{k-5}{3}+2+2$ other cities. So $\frac{k-5}{3}+k+5 \leq 2018$, so $4 k-5+15 \leq 3 \cdot 2018=6054$, so $4 k \leq 6044$, so $k \leq 1511$.
We conclude that the largest possible number of roads in a shortest route is 1511.