Can my calculator even handle this?

$9^{2004} \equiv (9^{\phi(100)})^{50}\cdot 9^4 \equiv 1 \cdot 6561 \equiv 61 \pmod{100}$

Hence, the sum of the tens and units place is $6+1= \boxed{7}$

As we can observe, it is can be expanded through Binomial Expansion so,

$9^{2004}$

$=$ $(10-1)^{2004}$

$=$ $10^{2004}$ - $\binom{2004}{1}$ $10^{2003}$ $+$ $...$ $+$ $\binom{2004}{2002}$ $10^2$ $-$ $\binom{2004}{2003}$ $10$ $+$ $1$

Using modulo 100,

-> $-$ $\binom{2004}{2003}$ $10$ $+$ $1$ mod 100

-> $\equiv$ $-40$ $+$ $1$ mod 100

-> $\equiv$ $-39$ mod 100

-> $\equiv$ $61$ mod 100

$=$ $6 + 1$ $=$ $7$

therefore, 7 is the sum.