Uniformly crossing a square table For every even number $n$ I can walk over $n\times n$  table starting from one (let's say bottom left) corner and finishing in the same corner so that every other field is visited exactly one time while bottom left corner is visited twice (one time from starting one when I finished). And if I continue walking on that route for many times I will have visited every field for the same number of times.
This is how to fill table with even $n$:

I am visiting every field exactly once. But is there any route that I can follow on $(2m+1) \times(2m+1)$ table (with odd $n$) so that if I continue following that route many times I get visited every field for the same number of times. This route doesn't need to visit every field exactly once I may visit twice or $k$ times. 
I didn't even get such a route for $3\times 3$ table and it seems to be impossible. Is there a way to find some route or any way to prove that it's impossible? 

If you consider a $(2m+1)\times (2m+1)$ checkerboard you'll see that you don't have the same amount of white and black cells. (One amount exceeds the other by one).
Any cyclic path on such a board needs to go through as many white as black cells (from a white cell you only can go to a black one and viceversa). So a cyclic hamiltonian path (one that goes through every cell exactly one and ends where it started) is not possible.
As a matter of fact, a cyclic path that doesn't repeat cells (other than the fact that the first and last cells are the same one) would have to go through an even amount of cells.  So a cyclic hamiltonian path over a  $(2m+1)\times (2k+1)$ grid is also impossible. 
