$N$ cells are marked on an $n\times n$ table so that at least one marked cel is among any four cells of the table which form the figure  [img]https://cdn.artofproblemsolving.com/attachments/2/2/090c32eb52df31eb81b9a86c63610e4d6531eb.png[/img] (tbe figure may be rotated). Find the smallest possible value of $N$.

(E. Barabanov)
To solve this problem, we need to find the smallest number of cells, \( N \), that must be marked on an \( n \times n \) table such that any set of four cells forming the given figure (a 2x2 square) contains at least one marked cell. 

1. **Coloring the Board:**
   Consider coloring the board like a chessboard, where each cell is either black or white in an alternating pattern. This means that each 2x2 square will contain two black cells and two white cells.

2. **Observation:**
   Notice that in any 2x2 square, there are always two black cells and two white cells. Therefore, if we mark all the black cells (or all the white cells), every 2x2 square will contain at least one marked cell.

3. **Counting the Marked Cells:**
   Since the board is \( n \times n \), there are \( n^2 \) cells in total. Half of these cells will be black and the other half will be white. Therefore, the number of black cells (or white cells) is:
   \[
   \left\lceil \frac{n^2}{2} \right\rceil
   \]
   This is because if \( n \) is even, \( n^2 \) is even, and if \( n \) is odd, \( n^2 \) is odd, and the ceiling function ensures we round up to the nearest whole number.

4. **Verification:**
   To verify, consider any 2x2 square on the board. Since we have marked all the black cells (or all the white cells), each 2x2 square will contain at least one marked cell, satisfying the problem's condition.

Therefore, the smallest possible value of \( N \) is:
\[
N = \left\lceil \frac{n^2}{2} \right\rceil
\]

The final answer is \( \boxed{\left\lceil \frac{n^2}{2} \right\rceil} \).