smc

Since N is a four digit number, assume WLOG that $N=1000a+100b+10c+d$, where a is the thousands digit, b is the hundreds digit, c is the tens digit, and d is the ones digit. Then, $\frac{1}{9}N=100b+10c+d$, so $N=900b+90c+9d$ Set these equal to each other: $$1000a+100b+10c+d=900b+90c+9d$$ $$1000a=800b+80c+8d$$ $$1000a=8(100b+10c+d)$$ Notice that $100b+10c+d=N-1000a$, thus: $$1000a=8(N-1000a)$$ $$1000a=8N-8000a$$ $$9000a=8N$$ $$N=1125a$$ Go back to our first equation, in which we set $N=1000a+100b+10c+d$, Then: $$1125a=1000a+100b+10c+d$$ $$125a=100b+10c+d$$ The upper limit for the right hand side (RHS) is $999$ (when $b=9$, $c=9$, and $d=9$). It's easy to prove that for an $a$ there is only one combination of $b,c,$ and $d$ that can make the equation equal. Just think about the RHS as a three digit number $bcd$. There's one and only one way to create every three digit number with a certain combination of digits. Thus, we test for how many as are in the domain set by the RHS. Since $125\cdot 7=875$ which is the largest $a$ value, then $a$ can be $1$ through $7$, giving us the answer of $\boxed{7}$ IronicNinja~