$k$ marbles are placed onto the cells of a $2024 \times 2024$ grid such that each cell has at most one marble and there are no two marbles are placed onto two neighboring cells (neighboring cells are defined as cells having an edge in common).

a) Assume that $k=2024$. Find a way to place the marbles satisfying the conditions above, such that moving any placed marble to any of its neighboring cells will give an arrangement that does not satisfy both the conditions.
b) Determine the largest value of $k$ such that for all arrangements of $k$ marbles satisfying the conditions above, we can move one of the placed marble onto one of its neighboring cells and the new arrangement satisfies the conditions above.
### Part (a)

1. **Arrange Marbles on the Diagonal:**
   - Place each of the 2024 marbles on one of the two main diagonals of the \(2024 \times 2024\) grid. For example, place a marble at each cell \((i, i)\) for \(i = 1, 2, \ldots, 2024\).

2. **Verify the Condition:**
   - Each marble is placed such that no two marbles are in neighboring cells. Neighboring cells share an edge, but in this arrangement, each marble is only diagonally adjacent to another marble, not sharing an edge.

3. **Check the Movement Condition:**
   - Moving any marble to a neighboring cell (either up, down, left, or right) will place it in a cell that is already occupied by another marble, violating the condition that no two marbles are in neighboring cells.

Thus, the arrangement satisfies the conditions, and moving any marble will violate the conditions.

### Part (b)

1. **Claim: \(k = 2023\) is the Largest Value:**
   - We need to show that for \(k = 2023\), there exists an arrangement where moving any marble to a neighboring cell will still satisfy the conditions.

2. **Assume \(k = 2024 - n\) Marbles:**
   - Let \(n \geq 1\). We need to show that there is always a way to move a marble if \(k < 2024\).

3. **Free Rows and Columns:**
   - Since there are fewer than 2024 marbles, there will be rows and columns without any marbles, called **free** rows and columns.
   - Let \(p\) be the number of free columns and \(q\) be the number of free rows. Without loss of generality, assume \(p \geq q\).

4. **Claim 1: No Two Free Rows are Adjacent:**
   - If there were two adjacent free rows, there must be a marble adjacent to this block. Moving this marble into the block would create a valid arrangement, contradicting the assumption.

5. **Claim 2: The Leftmost Column is Not Free:**
   - If the leftmost column were free, moving a marble from the next column to the leftmost column would create a valid arrangement, contradicting the assumption.

6. **Marbles in Non-Free Columns:**
   - For each free column, the column immediately to the left contains a marble. If a marble is at \((x, y)\), there must be another marble at \((x+2, y)\) to prevent moving the marble to \((x+1, y)\).

7. **Counting Marbles and Rows:**
   - There are \(p\) free columns, so there are at least \(p\) pairs of marbles with a distance vector \((2, 0)\). These pairs use up \(p\) rows.
   - The remaining \(2024 - n - 2p\) marbles can use at most \(2024 - n - 2p\) rows.
   - Therefore, at most \(p + (2024 - n - 2p) = 2024 - n - p\) rows have marbles, implying at least \(p + n\) rows are free.

8. **Contradiction:**
   - Since \(p \geq q\), the number of free rows \(q\) must be at least \(p + n\), which is a contradiction.

Thus, the largest value of \(k\) such that for all arrangements of \(k\) marbles, we can move one marble to a neighboring cell and still satisfy the conditions is \(k = 2023\).

The final answer is \( \boxed{ k = 2023 } \)