I know that $9^n$ has a pattern in its last two digits that repeats every 10 powers. For example, $9^1 = 09, 9^2 = 81, 9^3 = 29, 9^4 = 61, 9^5 = 49, 9^6 = 41, 9^7 = 69, 9^8 = 21, 9^9 = 89, 9^{10} = 01$, and then it starts over with $9^{11} = 09$. So, to find the last two digits of $9^{2004}$, I just need to figure out which power of 9 in the cycle it corresponds to. Since the cycle has length 10, I can divide 2004 by 10 and look at the remainder. 2004 divided by 10 is 200 with a remainder of 4. That means $9^{2004}$ has the same last two digits as $9^4$, which is 61. The tens digit is 6 and the units digit is 1, so the sum of the tens and units digits is 6 + 1 = 7. 
What is the sum of the tens digit and the units digit in the decimal representation of $9^{2004}$?