Take the number  $2^{2004}$ and calculate the sum  $S$ of all its digits. Then the sum of all the digits of  $S$ is calculated to obtain  $R$. Next, the sum of all the digits of $R$is calculated   and so on until a single digit number is reached. Find it.  (For example if we take   $2^7=128$, we find that   $S=11,R=2$. So in this case of   $2^7$ the searched digit will be $2$).
1. **Define the function \( q(N) \)**: The function \( q(N) \) represents the sum of the digits of \( N \). For example, if \( N = 128 \), then \( q(128) = 1 + 2 + 8 = 11 \).

2. **Understand the problem**: We need to repeatedly apply the function \( q \) to \( 2^{2004} \) until we reach a single digit number. This process can be simplified using properties of modular arithmetic.

3. **Use properties of digit sum and modular arithmetic**: It is known that the sum of the digits of a number \( N \) is congruent to \( N \mod 9 \). This is because:
   \[
   q(N) \equiv N \pmod{9}
   \]
   This property holds because the sum of the digits of a number is equivalent to the number itself modulo 9.

4. **Apply the property to \( 2^{2004} \)**: We need to find \( 2^{2004} \mod 9 \). To do this, we can use the fact that powers of 2 modulo 9 repeat in a cycle. Let's find the cycle:
   \[
   \begin{align*}
   2^1 &\equiv 2 \pmod{9}, \\
   2^2 &\equiv 4 \pmod{9}, \\
   2^3 &\equiv 8 \pmod{9}, \\
   2^4 &\equiv 16 \equiv 7 \pmod{9}, \\
   2^5 &\equiv 32 \equiv 5 \pmod{9}, \\
   2^6 &\equiv 64 \equiv 1 \pmod{9}.
   \end{align*}
   \]
   We see that \( 2^6 \equiv 1 \pmod{9} \). This means that every 6th power of 2 is congruent to 1 modulo 9.

5. **Simplify the exponent**: Since \( 2^{2004} \) can be written as \( (2^6)^{334} \times 2^0 \) (because \( 2004 = 6 \times 334 \)), we have:
   \[
   2^{2004} \equiv (2^6)^{334} \equiv 1^{334} \equiv 1 \pmod{9}.
   \]

6. **Conclude the single digit**: Since \( 2^{2004} \equiv 1 \pmod{9} \), the repeated application of the function \( q \) will eventually reduce \( 2^{2004} \) to the single digit 1.

\(\blacksquare\)

The final answer is \( \boxed{ 1 } \).