The checkered plane is divided into dominoes. What is the maximum $k{}$ for which it is always possible to choose a $100\times 100$ checkered square containing at least $k{}$ whole dominoes?

[i]Proposed by S. Berlov[/i]
1. **Generalize the Problem:**
   We start by generalizing the problem to a \(2N \times 2N\) checkered square instead of a \(100 \times 100\) square. We aim to show that such a square contains at least \(N(2N-1)\) whole dominoes.

2. **Boundary Piercing Argument:**
   To show that a \(2N \times 2N\) square contains at least \(N(2N-1)\) whole dominoes, we need to demonstrate that at most \(2N\) dominoes are pierced by its boundary. 

3. **Lattice Definition:**
   For \(i = 1, 2, \ldots, 2N\), let \(L_i\) denote the lattice of \(2N \times 2N\) squares, with one of the vertices at \((i, i)\).

4. **Domino-Density Definition:**
   For a set of dominoes \(D\) and a set of grid-lines \(\mathcal{L}\), define the domino-density \(\rho_D(\mathcal{L})\) as the ratio between the number of dominoes in \(D\) which are pierced by a line in \(\mathcal{L}\) and the total number of dominoes in \(D\).

5. **Contradiction Assumption:**
   Suppose, for the sake of contradiction, that the boundary of any \(2N \times 2N\) square of any lattice \(L_i\) pierces at least \(2N+1\) dominoes.

6. **Large Square Consideration:**
   Consider a large \(2Nt \times 2Nt\) square and let \(D\) be the set of dominoes that do not lie entirely outside this square. Since \(|D| = 2(Nt)^2 + O(t)\) and using our assumption, we get:
   \[
   1 = \sum_{i=1}^{2N} \rho_D(L_i) \geq \sum_{i=1}^{2N} \frac{1/2(t-1)^2(2N+1)}{2(Nt)^2 + O(t)} = \frac{N(2N+1)(t-1)^2}{2(Nt)^2 + O(t)}
   \]
   This inequality leads to a contradiction for large enough \(t\).

7. **Conclusion from Contradiction:**
   Therefore, there exists a \(2N \times 2N\) square whose boundary pierces at most \(2N\) dominoes.

8. **Optimality Argument:**
   To show that one cannot do better than this, consider the brick-tiling of the plane. In a brick-tiling, each \(2N \times 2N\) square will have exactly \(N(2N-1)\) whole dominoes, as the boundary will pierce exactly \(2N\) dominoes.

\(\blacksquare\)

The final answer is \( \boxed{ N(2N-1) } \)