Let $n{}$ be a positive integer. What is the smallest sum of digits that $5^n + 6^n + 2022^n$ can take?
1. **Evaluate the sum of digits for \( n = 1 \):**
   \[
   5^1 + 6^1 + 2022^1 = 5 + 6 + 2022 = 2033
   \]
   The sum of the digits of 2033 is:
   \[
   2 + 0 + 3 + 3 = 8
   \]

2. **Evaluate the sum of digits for \( n = 2 \):**
   \[
   5^2 + 6^2 + 2022^2 = 25 + 36 + 2022^2
   \]
   Since \( 2022^2 \) is a large number, we can approximate the sum of its digits. However, it is clear that the sum of the digits will be larger than 8.

3. **Consider \( n \geq 3 \):**
   - We need to analyze the expression \( 5^n + 6^n + 2022^n \) modulo 4 and 9.
   - First, consider modulo 4:
     \[
     5 \equiv 1 \pmod{4}, \quad 6 \equiv 2 \pmod{4}, \quad 2022 \equiv 2 \pmod{4}
     \]
     Therefore:
     \[
     5^n \equiv 1^n \equiv 1 \pmod{4}, \quad 6^n \equiv 2^n \pmod{4}, \quad 2022^n \equiv 2^n \pmod{4}
     \]
     Summing these:
     \[
     5^n + 6^n + 2022^n \equiv 1 + 2^n + 2^n \pmod{4}
     \]
     For \( n \geq 3 \), \( 2^n \equiv 0 \pmod{4} \):
     \[
     5^n + 6^n + 2022^n \equiv 1 + 0 + 0 \equiv 1 \pmod{4}
     \]

   - Next, consider modulo 9:
     \[
     5 \equiv 5 \pmod{9}, \quad 6 \equiv 6 \pmod{9}, \quad 2022 \equiv 6 \pmod{9}
     \]
     Therefore:
     \[
     5^n \pmod{9}, \quad 6^n \pmod{9}, \quad 2022^n \equiv 6^n \pmod{9}
     \]
     Summing these:
     \[
     5^n + 6^n + 2022^n \equiv 5^n + 6^n + 6^n \pmod{9}
     \]
     For \( n \geq 3 \), we need to check the periodicity of \( 5^n \pmod{9} \):
     \[
     5^1 \equiv 5 \pmod{9}, \quad 5^2 \equiv 25 \equiv 7 \pmod{9}, \quad 5^3 \equiv 125 \equiv 8 \pmod{9}, \quad 5^4 \equiv 625 \equiv 4 \pmod{9}
     \]
     The sequence repeats every 6:
     \[
     5^5 \equiv 20 \equiv 2 \pmod{9}, \quad 5^6 \equiv 10 \equiv 1 \pmod{9}
     \]
     Therefore:
     \[
     5^n + 6^n + 6^n \equiv 5^n + 2 \cdot 6^n \pmod{9}
     \]

4. **Analyze the last digit:**
   - The last digit of \( 5^n \) is always 5.
   - The last digit of \( 6^n \) is always 6.
   - The last digit of \( 2022^n \) is always 2.

   Therefore, the last digit of \( 5^n + 6^n + 2022^n \) is:
   \[
   5 + 6 + 2 = 13 \implies 3
   \]

5. **Conclusion:**
   - The smallest sum of digits for \( n = 1 \) is 8.
   - For \( n \geq 2 \), the sum of digits is larger than 8.

Therefore, the smallest possible sum of the digits of \( 5^n + 6^n + 2022^n \) is \( \boxed{8} \).