, 2013

Answer: for all $k \ge 2$.

Solution:
Such an arrangement is impossible for $k = 1$. In order to make all row sums and column sums even, both odd numbers should occur in the same row and also in the same column, which is impossible. In the rest, let $0$ and $1$ denote any even and odd number, respectively. For $k = 2$, one suitable arrangement is shown below:
$$\begin{matrix} 1 & 1 & 1 & 0 \\ 1 & 1 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{matrix}$$
We now show how to obtain a suitable arrangement for $k + 1$ from any suitable arrangement for $k$. Add $0$, $1$ to the end of the first $k - 1$ rows and add $1$, $0$ to the end of the following $k + 1$ rows. This fills the $2k \times 2$ strip appearing in the right end of the table. Fill the $2 \times 2k$ strip below the original part of the table similarly. Let the remaining $2 \times 2$ corner be $\begin{smallmatrix} 0 & 1 \\ 1 & 0 \end{smallmatrix}$. Then the parity of the sum of each old row and column is inverted. Each new column or row contains either $k$ or $k + 2$ odd numbers, whence the parity of the row and column sums is the opposite to that of $k + 1$. Hence the extended table meets the requirements.