Solution:

Group the terms four at a time. For $k = 1, 2, \ldots, 25,$ the block ending at $(4k)^2$ is $$(4k)^2 + (4k-1)^2 - (4k-2)^2 - (4k-3)^2 = 2(8k - 2) + 2(8k - 4) = 32k - 12,$$ using the difference of squares $a^2 - b^2 = (a + b)(a - b)$ with $a - b = 2$ twice.

Summing over $k = 1$ to $25,$ $$N = 32 \cdot \frac{25 \cdot 26}{2} - 12 \cdot 25 = 10400 - 300 = 10100,$$ so the remainder when $N$ is divided by $1000$ is $100.$