A positive integer $k$ is given. Initially, $N$ cells are marked on an infinite checkered plane. We say that the cross of a cell $A$ is the set of all cells lying in the same row or in the same column as $A$. By a turn, it is allowed to mark an unmarked cell $A$ if the cross of $A$ contains at least $k$ marked cells. It appears that every cell can be marked in a sequence of such turns. Determine the smallest possible value of $N$.
1. **Understanding the Problem:**
   We need to determine the smallest number of initially marked cells, \( N \), such that every cell on an infinite checkered plane can eventually be marked by a sequence of turns. A turn allows marking an unmarked cell \( A \) if the cross of \( A \) (all cells in the same row or column as \( A \)) contains at least \( k \) marked cells.

2. **Construction:**
   - Consider a \( k \times k \) isosceles right triangle on the checkered plane, with the hypotenuse colored.
   - Mark every cell within this triangle. This ensures that each row and each column within the triangle has a sufficient number of marked cells to propagate marking to the entire plane.
   - Specifically, the number of marked cells in each row and column increases as we move from the right angle to the hypotenuse.

3. **Counting the Marked Cells:**
   - The number of marked cells in the \( i \)-th row (or column) of the triangle is \( \lceil \frac{k-i+1}{2} \rceil \).
   - Summing these values from \( i = 1 \) to \( k \) gives the total number of initially marked cells:
     \[
     N = \sum_{i=1}^{k} \lceil \frac{i}{2} \rceil
     \]

4. **Bounding the Problem:**
   - To ensure that marking can start, at least one row or column must have \( \lceil \frac{k}{2} \rceil \) marked cells.
   - By induction, if we can solve the problem for \( k-1 \), we can solve it for \( k \) by adding the necessary cells to meet the \( k \)-requirement.

5. **Base Case:**
   - For \( k = 1 \), the problem is trivial as marking one cell is sufficient.

6. **Inductive Step:**
   - Assume the problem is solved for \( k-1 \). For \( k \), we need to ensure that at least one row or column has \( \lceil \frac{k}{2} \rceil \) marked cells. This can be achieved by adding the necessary cells to the configuration for \( k-1 \).

7. **Conclusion:**
   - The smallest number of initially marked cells \( N \) is given by:
     \[
     N = \lceil \frac{k}{2} \rceil + \lceil \frac{k-1}{2} \rceil + \cdots + \lceil \frac{1}{2} \rceil
     \]

The final answer is \( \boxed{ \lceil \frac{k}{2} \rceil + \lceil \frac{k-1}{2} \rceil + \cdots + \lceil \frac{1}{2} \rceil } \).