Solution:

Let $a$ be the two-digit number and $b$ the three-digit number. The condition is $1000a + b = 9ab,$ which rearranges to $b(9a - 1) = 1000a.$ Since $\gcd(9a - 1, a) = 1,$ the number $9a - 1$ must divide $1000.$

For a two-digit $a,$ $9a - 1$ runs from $89$ to $890,$ and $9a - 1 \equiv 8 \pmod 9.$ The only divisor of $1000$ in that range congruent to $8$ modulo $9$ is $125,$ giving $a = 14$ and $b = \frac{1000 \cdot 14}{125} = 112,$ which is indeed a three-digit number. Check: $14112 = 9 \cdot 14 \cdot 112.$

The requested sum is $14 + 112 = 126.$