points on the line $y=-x$ are given by
$n(n+1)+1,~0 \leq n$
a little toying shows that $(31)(32)+1=993$ is the largest one of these numbers 1000 or less.
so we have
$N = \sum \limits_{n=0}^{31} n(n+1)+1= 10416 + 496 + 31 = 10943$
I keep getting a different answer, off by 1, but I think I know why:
$\displaystyle \sum _{i=0} ^{31} 1= 32$, not 31, so final answer is 10944. What clued me in is that each of the diagonals is an odd number, so having an even number of entries on the diagonal would mean the sum must be even.