jmc

We know that the modular inverse exists because $11$ and $1000$ are relatively prime. Notice that $1000=10^3$ and that $11=10+1$. Since $11\cdot 11^{-1}\equiv 1(\mathrm{mod}1000)$, it follows that $(10+1)\cdot 11^{-1}=10^{3}k+1$ for some integer $k$. We recognize the potential sum of cubes factorization: if $k=1$, then $$10^3+1=1001=11\cdot (10^2-10+1)=11\cdot 91.$$Thus, $11^{-1}\equiv \boxed{91}(\mathrm{mod}1000)$.