jmc

We notice that both $A$ and $B$ are factors of 999,999. Specifically $$9A=999999$$and $$7B=999999.$$Taken modulo 1,000,000 these equations read $$\begin{array}{rl}9A& \equiv -1(\mathrm{mod}1,000,000)\\ 7B& \equiv -1(\mathrm{mod}1,000,000)\end{array}$$We are set if we multiply these equations: $$(9A)(7B)\equiv 1(\mathrm{mod}1,000,000)$$so $N=9\cdot 7=\boxed{63}$ is the multiplicative inverse to $AB$ modulo 1,000,000.