Answer:
.
Step-by-step explanation:
See below for a proof of why all but the first digit of this 
must be "
".
Taking that lemma as a fact, assume that there are 
digits in after the first digit, 
:
, where is a positive integer.
Sum of these digits:
.
Since is a digit, it must be an integer between 
and . The only possible value that would ensure is 
and 
.
Therefore:
.
.
.
Hence, the sum of the digits of 
would be 
.
Lemma: all digits of this other than the first digit must be "".
Proof:
The question assumes that 
is the smallest positive integer whose sum of digits is 
. Assume by contradiction that the claim is not true, such that at least one of the non-leading digits of is not "".
For example: 
, where , 
, 
, and 
are digits. (It is easy to show that contains at least 
digits.) Assume that 
is one of the non-leading non-"" digits.
Either of the following must be true:
- , the digit in front of is a "", or
- , the digit in front of is not a "".
If , the digit in front of , is a "", then let 
be with that "
" digit deleted: 
.
The digits of would still add up to :
.
However, with one fewer digit, 
. This observation would contradict the assumption that is the smallest positive integer whose digits add up to 
.
On the other hand, if , the digit in front of , is not "", then 
would still be a digit.
Since is not the digit , 
would also be a digit.
let be with digit replaced with , and replaced with : 
.
The digits of would still add up to :
.
However, with a smaller digit in place of , . This observation would also contradict the assumption that is the smallest positive integer whose digits add up to .
Either way, there would be a contradiction. Hence, the claim is verified: all digits of this other than the first digit must be "".
Therefore, would be in the form: 
, where , the leading digit, could also be .
