3. Given a positive integer $k$. On a grid plane, $N$ cells are pre-marked. The set of all cells in the same row or column as cell $A$ is called the "cross" of cell $A$. If in the cross of an unmarked cell $A$, $k$ other cells have already been marked, then cell $A$ can also be marked. It is known that after a series of such operations, all cells on the entire grid plane can be marked. For what minimum $N$ can this situation occur?
3. Let $N(k)$ denote the answer in the problem.

Let $f(k)=\left[\frac{k+1}{2}\right]\left[\frac{k+2}{2}\right]$ (where $[x]$ denotes the greatest integer not exceeding the real number $x$).
First, prove that for $k \geqslant 2$, we have
$$
N(k) \geqslant N(k-1)+\left[\frac{k+1}{2}\right].
$$

After marking the initial $N(k)$ cells, at least one more cell $A$ can still be marked. This indicates that either in the column of this cell or in the row of this cell, at least $\left[\frac{k+1}{2}\right]$ other cells have already been marked. For definiteness, assume this occurs in row $l$.

Imagine that all cells in row $l$ have been marked. Clearly, any cell can still be marked. Remove row $l$ from the grid and bring the two resulting half-planes together to form a new grid. Now, any cell can be marked on this new plane, and at each step, the cell is marked if its cross already contains $k-1$ marked cells (since one cell from this cross has been removed from row $l$). Therefore, initially, this new plane should have at least $N(k-1)$ marked cells. This indicates that in the original plane, at least $N(k-1)$ cells did not belong to row $l$; hence, inequality (1) follows.
Since $N(1)=1$, by the proven inequality (1), we have
$$
N(k) \geqslant \sum_{i=1}^{k}\left[\frac{i+1}{2}\right]=f(k).
$$

Next, we provide an inductive example: after marking $f(k)$ cells, any other cell on the plane can still be marked.

The figure shown by two ladders of heights $p=\left[\frac{k}{2}\right]$ and $q=\left[\frac{k+1}{2}\right]$ is precisely an example that meets the requirements, and it is not difficult to see that it contains exactly $f(k)$ cells.

When $k=1$, the conclusion is obvious: after marking one cell, any cell in its cross can certainly be marked, and then any other cell can be marked. To proceed with the induction, cells $a_{1}, a_{2}, \cdots, a_{p}$ can be marked consecutively. Thereafter, in their row, $p+q=k$ cells have already been marked, so any cell can be marked in it. This indicates that once this row is removed, the value of $k$ is reduced by 1, and the induction hypothesis can be applied to the remaining plane.