TST for BMO

Answer: For each $m$ coprime with $2024$.

Let us first assume that $(m, 2024) > 1$. Then the tourist, starting from the capital with a favourite number $r$, can reach only cities with numbers $km + r$ (mod $2024$), $k \in \mathbb{N}$. Since it is not a complete system of residues modulo $2024$, it means that the set of all favourite numbers consists of less than $2024$ numbers. Take now $2024$ cities and connect every two of them with a road. This is an example for which (i) can hold if all colours (favourite numbers) are used. That is, in this particular case it's impossible to reach every city starting from the capital.

Let us assume now that $(m, 2024) = 1$. Denote by $V$ the set of all cities and let $W \subset V$ be the maximal possible subset of vertices such that we can assign favourite numbers and starting from the capital $v_0$ we can reach every vertex in $W$. We prove $W = V$. Assume for the sake of contradiction $W \neq V$. Then, for any vertices $u \in W, v \in V \setminus W$ that are connected, we have
$$f(v) - f(u) \neq m \pmod{2024} \quad (1)$$
Now, we construct a different mapping $f_1$ by changing the assigning of vertices in $V \setminus W$. We set,
$$f_1(v) \in [1..2024], f_1(v) = f(v) - m \pmod{2024}, \forall v \in V \setminus W.$$
For any $v \in W$ we set $f_1(v) = f(v)$. Observe that under the new mapping $f_1$, condition (i) still holds. Indeed, if $f_1(u) = f_1(v)$ for some connected vertices $u \in W, v \in V \setminus W$, it implies $f(u) = f(v) + m$ and thus, we would reach $v$ from $u$ under the mapping $f$, which contradicts the maximality of $W$.

Therefore, the mapping $f_1$ satisfies (i), and moreover starting from $v_0$ we can reach any vertex in $W$. Since $W$ is maximal, it's not possible to access any vertex outside $W$. It allows us to change the assignment of favourite numbers again using the rule
$$f_2(v) = f_1(v) - m \pmod{2024}, \forall v \in V \setminus W$$
and so on. Suppose now, $u \in W, v \in V \setminus W$ are connected. There exists $k$ for which $f(u) + m = f(v) - km$, because $\{km : k = 1, 2, \dots, 2024\}$ is a complete system of residues modulo $2024$. But this means that under the mapping $f_k$, the vertex $v$ is accessible from $u$, which contradicts the maximality of $W$. $\square$