Suppose that $a_1, a_2, a_3, \ldots$ is an infinite geometric sequence such that for all $i \ge 1$, $a_i$ is a positive integer. Suppose furthermore that $a_{20} + a_{21} = 20^{21}$. If the minimum possible value of $a_1$ can be expressed as $2^a 5^b$ for positive integers $a$ and $b$, find $a + b$.

[i]Proposed by Andrew Wu[/i]
1. Let the common ratio of the geometric sequence be \( r \) and the first term be \( a_1 \). This means:
   \[
   a_{20} = a_1 \cdot r^{19} \quad \text{and} \quad a_{21} = a_1 \cdot r^{20}
   \]
   Given that \( a_{20} + a_{21} = 20^{21} \), we can write:
   \[
   a_1 \cdot r^{19} + a_1 \cdot r^{20} = 20^{21}
   \]
   Factoring out \( a_1 \cdot r^{19} \) from the left-hand side, we get:
   \[
   a_1 \cdot r^{19} (1 + r) = 20^{21}
   \]

2. Since \( 20^{21} = (2 \cdot 2 \cdot 5)^{21} = 2^{42} \cdot 5^{21} \), we have:
   \[
   a_1 \cdot r^{19} (1 + r) = 2^{42} \cdot 5^{21}
   \]

3. We need to find \( r \) such that \( r^{19} \) divides \( 2^{42} \cdot 5^{21} \). The possible values for \( r \) are the divisors of \( 2^{42} \cdot 5^{21} \), which are \( 1, 2, 4, 5, 10, 20 \).

4. Testing each possible value of \( r \):
   - For \( r = 1 \):
     \[
     a_1 \cdot 1^{19} (1 + 1) = 2^{42} \cdot 5^{21} \implies a_1 \cdot 2 = 2^{42} \cdot 5^{21} \implies a_1 = 2^{41} \cdot 5^{21}
     \]
     This value of \( a_1 \) is very large.

   - For \( r = 2 \):
     \[
     a_1 \cdot 2^{19} (1 + 2) = 2^{42} \cdot 5^{21} \implies a_1 \cdot 2^{19} \cdot 3 = 2^{42} \cdot 5^{21} \implies a_1 \cdot 3 = 2^{23} \cdot 5^{21}
     \]
     Since \( 3 \) does not divide \( 2^{23} \cdot 5^{21} \), \( r = 2 \) is not a valid solution.

   - For \( r = 4 \):
     \[
     a_1 \cdot 4^{19} (1 + 4) = 2^{42} \cdot 5^{21} \implies a_1 \cdot 2^{38} \cdot 5 = 2^{42} \cdot 5^{21} \implies a_1 = 2^{4} \cdot 5^{20}
     \]
     This value of \( a_1 \) is smaller and valid.

   - For \( r = 5 \):
     \[
     a_1 \cdot 5^{19} (1 + 5) = 2^{42} \cdot 5^{21} \implies a_1 \cdot 5^{19} \cdot 6 = 2^{42} \cdot 5^{21} \implies a_1 \cdot 6 = 2^{42} \cdot 5^{2}
     \]
     Since \( 6 \) does not divide \( 2^{42} \cdot 5^{2} \), \( r = 5 \) is not a valid solution.

   - For \( r = 10 \):
     \[
     a_1 \cdot 10^{19} (1 + 10) = 2^{42} \cdot 5^{21} \implies a_1 \cdot 2^{19} \cdot 5^{19} \cdot 11 = 2^{42} \cdot 5^{21} \implies a_1 \cdot 11 = 2^{23} \cdot 5^{2}
     \]
     Since \( 11 \) does not divide \( 2^{23} \cdot 5^{2} \), \( r = 10 \) is not a valid solution.

   - For \( r = 20 \):
     \[
     a_1 \cdot 20^{19} (1 + 20) = 2^{42} \cdot 5^{21} \implies a_1 \cdot 2^{38} \cdot 5^{19} \cdot 21 = 2^{42} \cdot 5^{21} \implies a_1 \cdot 21 = 2^{4} \cdot 5^{2}
     \]
     Since \( 21 \) does not divide \( 2^{4} \cdot 5^{2} \), \( r = 20 \) is not a valid solution.

5. The minimum possible value of \( a_1 \) is \( 2^4 \cdot 5^{20} \). Therefore, \( a = 4 \) and \( b = 20 \).

The final answer is \( a + b = 4 + 20 = \boxed{24} \).