There must be dozens of solutions!

We need $10^m \equiv 10^n \pmod{1001}$ , and hence $10^{n-m} \equiv 1$ modulo $7$ , $11$ and $13$ . Since $10^6$ is the smallest power of $10$ that is congruent to $1$ modulo each of $7\,,\,11\,,\,13$ , the answer is simply the number of pairs of non-negative integers $0 \le m < n \le 99$ such that $m \equiv n \pmod{6}$ .

For $j = 0,1,2,3$ there are $17$ integers between $0$ and $99$ that are congruent to $j$ modulo $6$ , and hence there are $\binom{17}{2}$ pairs of integers $0 \le m < n \le 99$ such that $m \equiv n \equiv j \pmod{6}$ .

For $j=4,5$ there are $16$ integers between $0$ and $99$ that are congruent to $j$ modulo $6$ , and hence there are $\binom{16}{2}$ pairs of integers $0 \le m < n \le 99$ such that $m \equiv n \equiv j \pmod{6}$ .

Hence there are $4\binom{17}{2} + 2\binom{16}{2} = \boxed{784}$ pairs of non-negative integers $0 \le m < n \le 99$ such that $10^m \equiv 10^n \pmod{1001}$ .