jmc

The prime factorization of $1001=7\times 11\times 13$. We have $7\times 11\times 13\times k=10^j-10^i=10^i(10^{j-i}-1)$. Since $\text{gcd}(10^i=2^i\times 5^i,7\times 11\times 13)=1$, we require that $1001=10^3+1|10^{j-i}-1$. From the factorization $10^6-1=(10^3+1)(10^3-1)$, we see that $j-i=6$ works; also, $a-b|a^n-b^n$ implies that $10^6-1|10^{6k}-1$, and so any $\boxed{j-i\equiv 0(\mathrm{mod}6)}$ will work. To show that no other possibilities work, suppose $j-i\equiv a(\mathrm{mod}6),1\le a\le 5$, and let $j-i-a=6k$. Then we can write $10^{j-i}-1=10^a(10^{6k}-1)+(10^a-1)$, and we can easily verify that $10^6-1\nmid 10^a-1$ for $1\le a\le 5$. If $j-i=6,j\le 99$, then we can have solutions of $10^6-10^0,10^7-10^1,\cdots ⟹94$ ways. If $j-i=12$, we can have the solutions of $10^{12}-10^0,\cdots ⟹94-6=88$, and so forth. Therefore, the answer is $94+88+82+\cdots +4⟹16\left(\frac{98}{2}\right)=\boxed{784}$.