Solution:
If $n$ is even, it is possible to construct the path for any $k$. Indeed, one can exhibit a closed path that passes once through each room, and by making $k$ loops one obtains the watchman's journey.
Let us assume for convenience that the museum is oriented along the cardinal points. One way to construct this path is as follows: the watchman starts from the room in the south-west corner and proceeds along the south side until reaching the south-east corner.

At this point he still has to cross the rooms of an $(n-1) \times n$ rectangle, which he traverses in a serpentine manner. He moves north for $n-1$ rooms (see example above for $n=4$), then he shifts west and goes back south for $n-2$ rooms. At this point he has crossed the two easternmost rows of rooms. If he shifts west and repeats the procedure, since $n$ is even, he crosses all the rows of rooms in pairs and can eventually return to the room from which he started.
If instead $n$ is odd, the problem has a solution only for $k=1$. In this case it is easy to find the path, which however does not return to the starting point, by making a serpentine in the manner described above. To see that in the other cases there is no solution, let us think of coloring the rooms white and black, as on a chessboard, with black in the corners. It is easy to see that in this way the black rooms are one more than the white ones. Consequently, if the watchman could make a path that passes $k$ times through each room, he should have visited $k$ black rooms more than white ones. However he can

move from one room only to another of a different color, so the difference between black and white rooms visited can be at most 1 during the journey, and for $k>1$ there is no solution.