Explanation:

- To find the least 4-digit number \(x\) that leaves a remainder of 1 when divided by 2, 3, 4, 5, 6, and 7, \(x-1\) must be a multiple of the least common multiple (LCM) of these numbers.

- Calculate the LCM:

- The LCM of 2, 3, 4, 5, 6, and 7 is \( \text{lcm}(2, 3, 4, 5, 6, 7) = 420 \).

- Therefore, \(x - 1 = 420k\) for some integer \(k\).

- Rewrite the equation as \(x = 420k + 1\).

- Since \(x\) is a 4-digit number between 2000 and 2500, solve:

- \(2000 \leq 420k + 1 \leq 2500\).

- This simplifies to finding \(k\) such that \(1999 \leq 420k \leq 2499\).

- Calculate \(k\):

- Minimum \(k = \lceil \frac{1999}{420} \rceil \approx 5\)

- Maximum \(k = \lfloor \frac{2499}{420} \rfloor \approx 5\)

- Use \(k = 5\), then calculate \(x\):

- \(x = 420 \times 5 + 1 = 2101\).

- The sum of the digits of 2101 is \(2 + 1 + 0 + 1 = 4\).

- Correct Answer: \(\textcolor{green}{\checkmark}\) 4