> Official solution:
> Scenario 1:  Move each tenant to the next higher room number.  In this
> case
> room 1 becomes free for La Poisson.
> Scenario 2:  Move each tenant to the room with double their room number
> (e.g. tenant of room n goes to room 2n).  In this case, the odd numbered
> rooms are free so Herr Banach can assign waiting customer k to room 2k-1
> (e.g. the first in line goes to room 1, the second to room 3, the third to
> room 5, etc.)

Leave it to a bunch of mathematicians to complicate things.  Both solutions
force everybody to move.  A much easier solution, at least for the guests,
is simply to renumber the rooms.