smc

The sum of the digits is at most $9+9+9=27$. Therefore the number is at most $6\cdot 27=162$. Out of the numbers $1$ to $162$ the one with the largest sum of digits is $99$, and the sum is $9+9=18$. Hence the sum of digits will be at most $18$. Also, each number with this property is divisible by $6$, therefore it is divisible by $3$, and thus also its sum of digits is divisible by $3$. Thus, the number is divisible by $18$. We only have six possibilities left for the sum of the digits: $3$, $6$, $9$, $12$, $15$, and $18$, but since the number is divisible by $18$, the digits can only add to $9$ or $18$. This leads to the integers $18$, $36$, $54$, $72$, $90$, and $108$ being possibilities. We can check to see that $\boxed{1}$ solution: the number $54$ is the only solution that satisfies the conditions in the problem.