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If $N$ is the desired number, then $N^{2}-N \equiv 0(\bmod 100)$. Also, $N^{2}-N=N(N-1)$, and $(N, N-1)=1$, so one of $N, N-1$ must be divisible by 8, and the other by 125. If $N$ is the three-digit number divisible by 125, then to make 8 | $N-1$, $N$ must be 625.

Similarly, if $N-1$ is the three-digit number divisible by 125, then only when $N-1=375$, i.e., $N=376$, does 8 | $N$ hold.

When $k \geqslant 2$, $N-1 \mid N^{k-1}-1$, so $N^{k}-N=N\left(N^{k-1}-1\right)$ is divisible by $N(N-1)=$ $N^{2}-N$. Therefore, if $N^{2}-N \equiv 0(\bmod 1000)$, then $N^{k}-N \equiv 0(\bmod 1000),(k \geqslant 2)$. Thus, $N=625,376$ are the solutions.