Solution:

The condition is that $1000 \mid x^{2}-x = x(x-1)$. Since $1000 = 2^{3} \cdot 5^{3}$, and $2$ cannot divide both $x$ and $x-1$, $2^{3} = 8$ must divide one of them. Similarly, $5^{3} = 125$ must divide either $x$ or $x-1$. We try successive values of $x$ that are congruent to $0$ or $1$ modulo $125$ and see which ones have the property that $x$ or $x-1$ is divisible by $8$. It is easy to check that $125, 126, 250, 251$, and $375$ do not work, but the next value, $376$, does, so this is the answer.