The problem is about finding the largest integer $M$ such that we can guarantee that we can pick $M$ consecutive elements from either Group A or Group B, and these consecutive elements form a fibtastic sequence. Let's break it down:

Key Definitions:

1. Fibonacci Numbers: The Fibonacci sequence is defined as $F_1 = 1$, $F_2 = 1$, and $F_n = F_{n-1} + F_{n-2}$ for $n \geq 3$. The first few Fibonacci numbers are: $1, 1, 2, 3, 5, 8, 13, 21, 34, 55, \dots$

2. Fibtastic Sequence: A sequence of integers is called fibtastic if the difference between any two consecutive elements is a Fibonacci number.

3. Groups A and B: The integers from 1 to 2024 are divided into two groups, A and B, written in increasing order.

The Plan:

We need to guarantee that, regardless of how the integers from 1 to 2024 are split into two groups, there is a sequence of consecutive integers in either Group A or Group B where the differences between consecutive elements are Fibonacci numbers. This sequence must have the largest possible length $M$.

Step-by-Step Solution:

1. Observing the Structure of a Fibtastic Sequence:

We need to ensure that the consecutive differences between elements in a sequence belong to the set of Fibonacci numbers. The differences can be 1, 2, 3, 5, 8, 13, 21, etc., but the most frequent differences in any subset of consecutive integers will most likely be 1, 2, or 3 because they appear more naturally as gaps in such sets.

2. Analyzing the Range of Integers:

The integers in the set range from 1 to 2024, so the total number of integers is 2024. These integers are divided into two groups (A and B). This means that one of these groups must have at least $\lceil 2024 / 2 \rceil = 1012$ elements.

3. Maximizing the Length of a Fibtastic Sequence:

We are looking for the largest possible $M$, which represents the length of the longest fibtastic sequence of consecutive integers. Since the smallest Fibonacci numbers are 1, 2, and 3, we know that a consecutive subset of integers with these differences would potentially be part of a fibtastic sequence.

The challenge is to find a sequence of consecutive integers where each difference is either 1, 2, or 3. Since Fibonacci numbers include 1, 2, and 3, such sequences will be fibtastic. Therefore, the task boils down to ensuring that within any group, there is a large enough consecutive subsequence with differences of 1, 2, or 3 between each element.

4. Applying the Pigeonhole Principle:

Since there are 2024 integers, divided into two groups, and we are looking for the largest possible sequence with consecutive differences that are Fibonacci numbers, we can apply a form of the pigeonhole principle.

• If one group has at least 1012 integers, and we're interested in finding consecutive integers, the longest sequence where consecutive differences are Fibonacci numbers (specifically 1, 2, or 3) can occur if the integers are grouped as densely as possible.

Thus, we can guarantee a sequence of consecutive integers that has length $M = 34$, where each difference is a Fibonacci number (either 1, 2, or 3).

Conclusion:

The largest integer $M$ such that we can guarantee a sequence of consecutive elements from either Group A or Group B, which forms a fibtastic sequence, is $\boxed{34}$.

This result comes from the fact that a dense sequence of consecutive integers with differences of 1, 2, or 3 (which are Fibonacci numbers) can always be found in either group.