The plate changing can not go on infinitely. Translating the problem to graph we'd have :We have a directed graph with vertices{1,2,...,n} which has no directed cycles. Each time we pick an edge that goes from a vertex with a greater number to another vertex with lesser number and exchange their numbers. We are willing to prove that this number exchanging ends after finite moves. First we prove that number n changes its place finite times. Since n is the largest number each time it goes forward on a path. Since we have no directed cycles n keeps on moving forward on a path so it will reach the end of it and won't move any further .Obviously the length of paths in our graph is finite. Thus after finite moves n stops. Similarly after n has stopped n-1 stops as well after finite moves and so on all the way down to 1.Therefore only finite number exchanging is possible and our claim is proved.

The answer is $\boxed{\text{no}}$. Consider each of the bus stops as labelled vertices in a graph and consider the bus routes to be a directed edges. It is clear that any higher-labelled vertex will swap labels if there is a directed edge going out to a lower-labelled vertex. Since there are no cycles within this graph, $N$ can only move a finite number of times. Once it ceases to move, we "remove" it from the graph and we are left with the case of $N-1$ vertices. This will continue downwards until we are left with $1$, at which point the labels will stop swapping.

The answer is no. Here is a brief sketch Consider obvious directed graph interpretation with vertex set $V$. Call a vertex a root if it has in-degree $0$ and we say a vertex $v$ is rooted at some root $r$ if there is a direct path from $r$ to $v$. Finally, let $S(v)$ denote the sum of the lengths of the paths from the roots of $v$ to $v$. We have the following observation, if $u \rightarrow v$ then $S(u)<S(v)$. If $n_v$ denotes the plate assigned to $v$ then we make the following hilarious observation. The ordinal \[ \lambda = \sum_{v \in V} n_v \cdot \omega^{S(v)} \]strictly decreases with every swapping of the plates. Since there are no infinite descending chains in the ordinals, the conclusion follows.

Posting for storage(My solution is same as the above's) The answer is NO Here is a sketch Assume by induction that for N-1 vertices it is finite Then prove that, for N it is finite(The proof is easy) We are done

No, it must end. FTSOC assume that it goes on forever. So some configuration appears twice. Now let the larger element of the first swap be denoted as $m$. So $m$ returns to its initial position after some set of moves. But this completes a cycle from the initial position of $m$ back to itself and it is given we cannot go from one city back to itself using the bus routes which is a contradiction. Thus we are done.