Let $q(n)$ denote the sum of the digits of a natural number $n$. Determine $q(q(q(2000^{2000})))$.
1. **Simplify the problem using properties of digit sums and modular arithmetic:**
   We need to determine \( q(q(q(2000^{2000}))) \). Notice that the sum of the digits of a number \( n \) modulo 9 is congruent to \( n \) modulo 9. Therefore, we can simplify the problem by considering \( 2000^{2000} \mod 9 \).

2. **Calculate \( 2000 \mod 9 \):**
   \[
   2000 = 2 + 0 + 0 + 0 = 2 \implies 2000 \equiv 2 \pmod{9}
   \]
   Thus, \( 2000^{2000} \equiv 2^{2000} \pmod{9} \).

3. **Use Euler's theorem to simplify \( 2^{2000} \mod 9 \):**
   Euler's theorem states that if \( \gcd(a, n) = 1 \), then \( a^{\phi(n)} \equiv 1 \pmod{n} \), where \( \phi \) is the Euler's totient function. For \( n = 9 \), \( \phi(9) = 6 \).

   Since \( \gcd(2, 9) = 1 \), we have:
   \[
   2^6 \equiv 1 \pmod{9}
   \]
   Therefore,
   \[
   2^{2000} = 2^{6 \cdot 333 + 2} = (2^6)^{333} \cdot 2^2 \equiv 1^{333} \cdot 2^2 \equiv 4 \pmod{9}
   \]

4. **Determine the sum of the digits of \( 2^{2000} \):**
   Since \( 2^{2000} \equiv 4 \pmod{9} \), the sum of the digits of \( 2^{2000} \) is congruent to 4 modulo 9. Therefore, \( q(2^{2000}) \equiv 4 \pmod{9} \).

5. **Apply the digit sum function iteratively:**
   Since \( q(2^{2000}) \equiv 4 \pmod{9} \), we have:
   \[
   q(q(2^{2000})) \equiv q(4) \equiv 4 \pmod{9}
   \]
   and
   \[
   q(q(q(2^{2000}))) \equiv q(4) \equiv 4 \pmod{9}
   \]

6. **Conclude the final result:**
   After applying the digit sum function three times, we still get a result congruent to 4 modulo 9. Since the digit sum function reduces the number to a single digit, the final result is 4.

The final answer is \(\boxed{4}\).