A child is playing coloring his chessboard and will paint each square either completely blue or completely red. To give it a personalized touch, he wants to paint the same number of red squares as blue squares, but he doesn't want to have two columns with the same number of red squares painted, nor does he want to have two rows with the same number of red squares painted. Can he manage to paint it while fulfilling these conditions? And what if instead of a normal 8x8 chessboard it were a giant 1000x1000 chessboard?
my effort so far is: if they have to be the same amount of painted for each color then
8x8 = 64 half is 32, so 32 red and 32 blue
as each row has to have a different amount we can start from 0 to 7, or go from 1 to 8 (the order doesn't matter)
If we do the sum painting that way it is:
possibility 1) 0 + ... + 7 = 28 -> is less than 32
possibility 2) 1 + ... + 8 = 36 -> is greater than 32
that is to say, I cannot, respecting the conditions, paint exactly half of one color and the other half of the other
Is there any other way to test this? does it work for the case 1000x1000? Can what I say be a demonstration or is it very dependent on the data of the problem?
Your argument works for any $n\times n$ square especially if you incorporate Gerry Myerson's comment.
This has $n^2$ little squares and so you need half of these, i.e. $\frac12n^2$, to be red.
If you do not have any rows with $0$ reds then the total number of reds in rows - each row with a distinct number so $1+2+\cdots+n$ - will be $\frac12n(n+1)$ which is too high.
If you do have a row with $0$ reds then you will not have any column with $n$ reds and the total number of reds in columns - each column with a distinct number so $0+1+\cdots+(n-1)$ - will be $\frac12n(n-1)$ which is too low.
So it is impossible.