jmc

$$(x-y)(x+y)=2000^2=2^8\cdot 5^6$$ Note that $(x-y)$ and $(x+y)$ have the same parities, so both must be even. We first give a factor of $2$ to both $(x-y)$ and $(x+y)$. We have $2^6\cdot 5^6$ left. Since there are $7\cdot 7=49$ factors of $2^6\cdot 5^6$, and since both $x$ and $y$ can be negative, this gives us $49\cdot 2=\boxed{98}$ lattice points.