Q:

What is the 1000th positive integer whose digit sum is equal to 9? 

1 - 9
2 - 18
3 - 27
...
1000 - ? 

A:

We can begin by estimating the number of digits the number we want has. 
If we want a number with at most $k$ digits to have digit sum $9$, this is the same as writing $9$ as an ordered sum of $k$ nonnegative integers.  By the stars and bars method (Wikipedia link) there are $\binom{9+k-1}{k-1}$ ways to do so.  
So there are $\binom{9+2-1}{2-1} = 10$ such numbers that are less than $100$, $\binom{9+3-1}{3-1} = 55$ such numbers less than $1000$, $\binom{9+4-1}{4-1} = 220$ such numbers less than $10000$, $\binom{9+5-1}{5-1} = 715$ such numbers less than $10^5$, and $\binom{9+6-1}{6-1}=2002$ such numbers less than $10^6$. So the $1000^{\text{th}}$ number in the sequence is between $10^5$ and $10^6$: it has $6$ digits. 
Next, we can figure out what the first digit is.  In addition to the $715$ numbers we've found with $5$ or fewer digits, the number of $6$-digit numbers whose first digit is $1$ in the sequence is $\binom{8+5-1}{5-1} = 495$, because its remaining $5$ digits sum to $8$, and $715+495 > 1000$. So the number we want is $1xxxxx$.
Similarly, we can figure out the second digit.  In addition to the $715$ numbers we've found with $5$ or fewer digits, the number of $6$-digit numbers beginning $10\dots$ in the sequence is $\binom{8+4-1}{4-1} = 165$ (there are $4$ digits remaining, that add up to $8$).  $715+165 = 880 < 1000$, so our number is past that.  The number of $6$-digit numbers beginning $11\dots$ in the sequence is $\binom{7+4-1}{4-1} = 120$, and $120 + 880$ puts at $1000$ exactly. 
This tells us that the number we want is the largest number of the form $11xxxx$ whose digits add up to $9$. We get the largest number by putting as large a digit as we can as early as we can, so the number we want is $117000$.

