Find the least positive integer $k$ such that when $\frac{k}{2023}$ is written in simplest form, the sum of the numerator and denominator is divisible by $7$.

[i]Proposed byMuztaba Syed[/i]
1. We need to find the least positive integer \( k \) such that when \(\frac{k}{2023}\) is written in simplest form, the sum of the numerator and denominator is divisible by \( 7 \).

2. First, note that \( 2023 \) is not divisible by \( 7 \). We can verify this by checking the divisibility rule for \( 7 \):
   \[
   2023 \div 7 = 289 \quad \text{(not an integer)}
   \]
   Therefore, \( 2023 \) is not divisible by \( 7 \).

3. Let \( \frac{k}{2023} \) be written in simplest form as \( \frac{a}{b} \). Since \( 2023 \) is not divisible by \( 7 \), \( b \) must be \( 2023 \) and \( a \) must be \( k \) (since \( k \) and \( 2023 \) are coprime).

4. We need the sum of the numerator and denominator to be divisible by \( 7 \):
   \[
   k + 2023 \equiv 0 \pmod{7}
   \]
   Simplifying, we find:
   \[
   k \equiv -2023 \pmod{7}
   \]

5. Calculate \( 2023 \mod 7 \):
   \[
   2023 \div 7 = 289 \quad \text{(remainder 0)}
   \]
   Therefore:
   \[
   2023 \equiv 0 \pmod{7}
   \]
   So:
   \[
   k \equiv 0 \pmod{7}
   \]

6. The smallest positive integer \( k \) that satisfies this congruence is \( k = 7 \).

7. Verify that \( \frac{7}{2023} \) is in simplest form:
   - The greatest common divisor (gcd) of \( 7 \) and \( 2023 \) is \( 1 \) (since \( 2023 \) is not divisible by \( 7 \)).

8. Check the sum:
   \[
   7 + 2023 = 2030
   \]
   Verify divisibility by \( 7 \):
   \[
   2030 \div 7 = 290 \quad \text{(an integer)}
   \]

Thus, the least positive integer \( k \) such that the sum of the numerator and denominator of \(\frac{k}{2023}\) in simplest form is divisible by \( 7 \) is \( k = 7 \).

The final answer is \( \boxed{7} \).