A sum with nines!

Since 1000 does not has a sum of digits of 9, we need only to consider from 100 to 999, 3-digit integers. Let the required 3-digit numbers be $n=\overline{abc} = 100a+ b + c$ , where $a$ , $b$ , and $c$ are $\mathbb Z \in [0,9]$ . Then the sum of $n$ is given by:

$\begin{aligned} S & = \sum_{a=1}^9 \sum_{b=0}^{9-a} \left(100a + 10 b + \color{#3D99F6} 9 - a - b \right) & \small \color{#3D99F6} \text{Note that }c = 9 - a - b, \\ & = \sum_{a=1}^9 \sum_{b=0}^{9-a} \left(99a + 9b + 9 \right) & \small \color{#3D99F6} \text{one }c \text{ for one }a \text{ and one }b. \\ & = 9 \sum_{a=1}^9 \sum_{b=0}^{9-a} \left(11a + b + 1 \right) \\ & = 9 \sum_{a=1}^9 \left((11a +1)(10-a) + \frac {(9-a)(10-a)}2 \right) \\ & = \frac 92 \sum_{a=1}^9 \left(110 + 199a - 21a^2 \right) \\ & = \frac 92 \left(110\cdot 9 + \frac {199\cdot 9 \cdot 10}2 - \frac {21\cdot 9 \cdot 10 \cdot 19}6 \right) \\ & = \boxed {17820} \end{aligned}$

$108+117+126+135+144+153+162+171+180+207+216+225+234+243+252+261+270+306+315+324+333+342+351+360+405+414+423+432+441+450+504+513+522+531+540+603+612+621+630+702+711+720+801+810+900=17820$

Ok let see. Now since thw sum of digits is 9 ,so it must be
a mutiple of 9.Now all multiples of 9 between 100 and 1000 do not have the sum nine ,some have 18 and only 1 has the sum 27(it is 999)In fact there are only 45 numbers with the sum of their digits equal to 9 between 100 to 1000.

Lets the digits be x,y,z Now x+y+z=9 . Z=9-(x+y) The no. Is of the form 100x+10y+z From two equations we get no is of the form, 9(11x+y+1) lets do case work , Case 1 when x=1,y is any no between 0 to 8. Case 2 when x=2,y is any no. Between 0 to 7. Case 3 when x=3, y os any no. Between 0 to 6. and so on... So between 100 to 200 we have nos 9×11,9×12,........,9×20. Between 200 to 300 we have no.s 9×23,9×24.....,9×30 and so on.... So between 100 and 200 numbers whose digits DO NOT sum up to nine are-9×21,9×22 Between 200 and 300 such numbers are-9×31,9×32,9×33 and so on... We get a symmetry. So between 900 and 1000 such no.s are- 9×101,9×102,........,9×110. But there is one more no that is 9×111=999 Its digits sum upto 27. Now we can sum up all such number we get-9(21+22+31+32+33+41+42+43+44+.....+101+102+103+104+105+106+107+108+109+110+111)=37530 sum all multiples of 9 between 100 and1000 which is=(100(216+99×9))÷2=55350 Subtracting this result from the previous reult we get=55350-37530=17820