, 2023

Solution:
The answer is $m = \left\lceil \frac{n}{4k} \right\rceil$.

Call a collection of cities independent if no two cities in the collection are joined by a road. Let $r$ and $k$ be integers such that $n = 4kq + r$ where $1 \leq r \leq 4k$.

First we show that $m \leq \left\lceil \frac{n}{4k} \right\rceil = q + 1$. Let $K_{i}$ denote a set of $i$ cities such that every pair of cities in $K_{i}$ is linked by a road. Consider a country containing $q$ copies of $K_{4k}$ and one copy of $K_{r}$. An independent set of cities in this country contains at most one city from each $K_{4k}$ or $K_{r}$ and therefore contains at most $q + 1$ cities. Now note that any partition of the cities of the country into two new countries partitions each $K_{4k}$ and $K_{r}$ into two sets. If $K_{i}$ where $i \leq 4k$ is partitioned into two sets of cities of sizes $a$ and $b$, then the number of roads between the two sets is $ab \leq \frac{(a + b)^2}{4} \leq k i$. Summing this inequality over all copies of $K_{4k}$ and $K_{r}$ yields that there are at most $k n$ roads between the two new countries. This implies that this particular country satisfies the given condition and it follows that $m \leq \left\lceil \frac{n}{4k} \right\rceil$.

Now we show that any country satisfying the given condition has an independent set containing at least $\left\lceil \frac{n}{4k} \right\rceil$ cities. Call a set of cities $i$-separable if it can be partitioned into $i$ disjoint independent sets of cities. Given a country satisfying the conditions, let $S$ be a largest set of cities in the country that is $2k$-separable. We prove that $|S| \geq n/2$. By definition of $S$, there exists a partition $A_{1}, A_{2}, \ldots, A_{2k}$ of the cities in $S$ such that each $A_{i}$ is independent. Let $|S| = t$. Assume for contradiction that $t < \frac{n}{2}$. There are at most $k n$ roads between $S$ and the rest of the country, which by the pigeonhole principle implies that there is a city $u$ not in $S$ that is connected to at most $\frac{k n}{n - t} < 2k$ cities by road. Therefore $u$ is joined by a road to at most $2k - 1$ cities in $S$, and there must be an independent subset $A_{i}$ such that $u$ is not linked by a road to any city in $A_{i}$. Adding $u$ to $S$ maintains the fact that $S$ is $2k$-separable but contradicts its maximality. Therefore it must follow that $t \geq \frac{n}{2}$. By the pigeonhole principle, one of the sets $A_{1}, A_{2}, \ldots, A_{2k}$ must contain at least $\frac{t}{2k} \geq \frac{n}{4k}$ cities. This proves the claim and therefore $m = \left\lceil \frac{n}{4k} \right\rceil$.