Problem 188

To solve the problem, let's break down the solution step by step, closely following the original solution provided:

1. Identify the Position of 2024 in the Sequence of Even Numbers:
- Each positive even number can be expressed as $2k$ for some positive integer $k$.
- Since $2024 = 2 \times 1012$, it means $2024$ is the $1012$-th even number in the sequence of all even numbers.

2. **Determine the Group Number ($n$) for 2024:**
- The sum of the first $n$ positive integers is given by the formula $\frac{n(n + 1)}{2}$.
- The groups are formed by consecutive even numbers, with the $1$st group having $1$ number, the $2$nd group having $2$ numbers, and so on.
- We need to find the group ($n$) such that the total count of numbers up to that group is just before or equal to $1012$.
- For $n = 44$, the total count is $\frac{(1 + 44) \times 44}{2} = 990$.
- For $n = 45$, the total count is $\frac{(1 + 45) \times 45}{2} = 1035$.
- Therefore, $2024$ falls into the $45$-th group because $990 < 1012 < 1035$.

3. **Determine the Position ($m$) of 2024 Within Its Group:**
- To find $m$, we calculate how far $2024$ is from the start of its group.
- The difference between $1012$ and the total count up to the previous group ($44$-th) is $1012 - 990 = 22$.
- This means $2024$ is the $22$-nd number in the $45$-th group.

4. **Calculate $m + n$:**
- With $m = 22$ (position in the group) and $n = 45$ (group number), we find $m + n = 45 + 22 = 67$.

Therefore, the final answer, following the rules and format specified, is $\boxed{67}$.