, 2006

i) After two steps: one can put into each cell of a small board of size $(4 \times 2)$ a marble. One can partition the given board of size $(2004 \times 2006)$ into small boards of size $(4 \times 2)$. Therefore, after some steps, one can put marbles into all cells of the given board so that the number of marbles in each cell is the same for all cells.

ii) We now prove by contradiction that in the second case, the response to the problem is "no". Indeed, suppose that the contrary: after some steps, there would be $k$ marbles ($k > 0$) in each cell of the given board of size $(2005 \times 2006)$. Color black all cells belonging to the odd rows and consider each non colored cell as white. Then, the number of black cells is equal to $1003 \times 2006$ and the number of white cells is equal to $1002 \times 2006$. But at each step we put exactly $2$ marbles into black cells and $2$ marbles in white cells. Therefore, after an arbitrary number of steps, the number of marbles in all black cells must be equal to the number of the marbles in all white cells. Consequently, we would have $1003 \times 2006 \times k = 1002 \times 2006 \times k$ and so get $1 = 0$. This contradiction proves our assertion.