Solution:

Answer no

Suppose it can be done. Divide the board into 4 quadrants. Suppose there are $b$ black and $995^2 - b$ white squares in the top left quadrant. Then there are $995^2 - b$ black and $b$ white squares in the bottom right quadrant (by the symmetry property).

If half the squares in each row are black, then half the squares in the first $995$ rows are black, so the number of black squares in the top right quadrant is $995 \cdot 1990 / 2 - b = 995^2 - b$. So if half the squares in each column are black, then half the squares in the right-hand half of the board are black, so $(995^2 - b) + (995^2 - b) = 995^2$, in other words, $b = 995^2 / 2$, which is impossible.