We will show that $m = 3$ for every $n > 2$.

Firstly we show that $m \ge 3$. It is clear that $m \ge 2$, so suppose that $m = 2$. We have $n$ cities, and the possible numbers for the sum of numbers assigned to roads going out from the given city are $n-1, n, n+1, \dots, 2n-2$ (the smallest possible sum is $1+1+\dots+1 = n-1$ and the largest is $2+2+\dots+2 = 2(n-1)$). Since there are exactly $n$ of them, every number is ID for some city. But it means that there is a city with every road having a number $1$ and a city with every road having a number $2$. But this is a contradiction since these cities are connected.

Now we will show that we can enumerate the roads with numbers $1, 2, 3$ such that the condition will be satisfied. Suppose that $n = 2k$. We will construct our enumeration in steps. To all roads from the first city assign the number $1$. Do the same with the second but assign the number $2$ to the road going to the $(2k-1)$-th city. For the third city assign $2$ to $(2k-1)$, $(2k-2)$-th cities. Continue this operation to the $k$-th city. Then the situation is: cities $1, 2, \ldots, k$ have the ID $2k-1, 2k, 2k+1, \ldots, 3k-2$ respectively (and we have finished assigning the number to their roads) and the cities $k+1, k+2, \ldots, 2k-1$ have the ID's $k, k+1, \ldots, 2k-2, 2k-1$ respectively. So now it suffices to assign number $3$ to every road between cities $k+1, k+2, \ldots, 2k-1$ and it is easy to see that all ID's are different. For the odd $n$ we use a very similar argument.