Let $a$ and $b$ be positive integers such that $(a^3 - a^2 + 1)(b^3 - b^2 + 2) = 2020$. Find $10a + b$.
1. We start with the given equation:
   \[
   (a^3 - a^2 + 1)(b^3 - b^2 + 2) = 2020
   \]

2. Factorize 2020:
   \[
   2020 = 2^2 \times 5 \times 101
   \]

3. We need to find values of \(a\) and \(b\) such that:
   \[
   (a^3 - a^2 + 1)(b^3 - b^2 + 2) = 4 \times 5 \times 101
   \]

4. Let's consider the factors of 2020. We observe that:
   \[
   b^3 - b^2 + 2 = 4 \quad \text{or} \quad b^3 - b^2 + 2 = 5
   \]

5. If \(b^3 - b^2 + 2 = 4\):
   \[
   b^3 - b^2 + 2 = 4 \implies b^3 - b^2 = 2 \implies b(b^2 - b) = 2
   \]
   This equation does not have integer solutions for \(b\).

6. If \(b^3 - b^2 + 2 = 5\):
   \[
   b^3 - b^2 + 2 = 5 \implies b^3 - b^2 = 3 \implies b(b^2 - b) = 3
   \]
   This equation does not have integer solutions for \(b\).

7. If \(b^3 - b^2 + 2 = 101\):
   \[
   b^3 - b^2 + 2 = 101 \implies b^3 - b^2 = 99 \implies b(b^2 - b) = 99
   \]
   This equation does not have integer solutions for \(b\).

8. If \(b^3 - b^2 + 2 = 2020\):
   \[
   b^3 - b^2 + 2 = 2020 \implies b^3 - b^2 = 2018 \implies b(b^2 - b) = 2018
   \]
   This equation does not have integer solutions for \(b\).

9. If \(b^3 - b^2 + 2 = 2\):
   \[
   b^3 - b^2 + 2 = 2 \implies b^3 - b^2 = 0 \implies b(b^2 - b) = 0
   \]
   This equation has the solution \(b = 1\).

10. If \(b = 1\):
    \[
    b^3 - b^2 + 2 = 1 - 1 + 2 = 2
    \]
    Then:
    \[
    (a^3 - a^2 + 1) \times 2 = 2020 \implies a^3 - a^2 + 1 = 1010
    \]

11. We need to solve:
    \[
    a^3 - a^2 + 1 = 1010
    \]
    Testing small values of \(a\):
    \[
    a = 10 \implies 10^3 - 10^2 + 1 = 1000 - 100 + 1 = 901 \quad (\text{not } 1010)
    \]
    \[
    a = 11 \implies 11^3 - 11^2 + 1 = 1331 - 121 + 1 = 1211 \quad (\text{not } 1010)
    \]
    \[
    a = 12 \implies 12^3 - 12^2 + 1 = 1728 - 144 + 1 = 1585 \quad (\text{not } 1010)
    \]

12. We observe that if \(b = 3\):
    \[
    b^3 - b^2 + 2 = 3^3 - 3^2 + 2 = 27 - 9 + 2 = 20 \quad (\text{not a factor of } 2020)
    \]

13. We need to re-evaluate the factorization approach. Let's try:
    \[
    b = 3 \implies b^3 - b^2 + 2 = 27 - 9 + 2 = 20
    \]
    Then:
    \[
    (a^3 - a^2 + 1) \times 20 = 2020 \implies a^3 - a^2 + 1 = 101
    \]

14. Solving:
    \[
    a^3 - a^2 + 1 = 101
    \]
    Testing small values of \(a\):
    \[
    a = 5 \implies 5^3 - 5^2 + 1 = 125 - 25 + 1 = 101
    \]

15. Therefore, \(a = 5\) and \(b = 3\).

16. Finally, calculate \(10a + b\):
    \[
    10a + b = 10 \times 5 + 3 = 50 + 3 = 53
    \]

The final answer is \(\boxed{53}\).