Among $18$ consecutive positive $3$-digit numbers, there exists $2$ numbers that is divisible by $9$. So the sum of its digits can either be $9$ or $27$. We see that there is only one $3$-digit number $999$ that has sum of it digits is $27$, so the remain number will have sum of it digits is $9$ and that number is also divisible by $9$. We are done.

shinichiman wrote: Among $18$ consecutive positive $3$-digit numbers, there exists $2$ numbers that is divisible by $9$. So the sum of its digits can either be $9$ or $27$. We see that there is only one $3$-digit number $999$ that has sum of it digits is $27$, so the remain number will have sum of it digits is $9$ and that number is also divisible by $9$. We are done. That's not complete enough. The sum of the digits could be 18 also. But then taking mod 2 gives the result.

mathguy5041 wrote: That's not complete enough. The sum of the digits could be 18 also. But then taking mod 2 gives the result. Thanks, I forgot $18$ is also divisible by $9$.

Among $18$ consecutive positive integers,there exists $a$ s.t. $18|a$.Since $9|a$,then $9|S(a)$.Since $S(a)\le 9\cdot 3=27$,then $S(a)=9,18,27$.If $S(a)=9,18$,then $S(a)|a$.If $S(a)=27$,then $a=999$ and $27|999$.Thus $S(a)|a$.Therefore the proof is completed.$\blacksquare$

In this problem,$18$ is the best evaluation. Because among $17$ numbers $559$~$575$,there is no one that is divisible by the sum of its digits.