jmc

We have that $N^2-N=N(N-1)\equiv 0\mathrm{mod}10000$ Thus, $N(N-1)$ must be divisible by both $5^4$ and $2^4$. Note, however, that if either $N$ or $N-1$ has both a $5$ and a $2$ in its factorization, the other must end in either $1$ or $9$, which is impossible for a number that is divisible by either $2$ or $5$. Thus, one of them is divisible by $2^4=16$, and the other is divisible by $5^4=625$. Noting that $625\equiv 1\mathrm{mod}16$, we see that $625$ would work for $N$, except the thousands digit is $0$. The other possibility is that $N$ is a multiple of $16$ and $N-1$ is a multiple of $625$. In order for this to happen,$$N-1\equiv -1(\mathrm{mod}16).$$Since $625\equiv 1(\mathrm{mod}16)$, we know that $15\cdot 625=9375\equiv 15\equiv -1\mathrm{mod}16$. Thus, $N-1=9375$, so $N=9376$, and our answer is $\boxed{937}$.