Let $m$ and $n$ be natural numbers with $mn$ even. Jetze is going to cover an $m \times n$ board (consisting of $m$ rows and $n$ columns) with dominoes, such that each domino covers exactly two squares, dominoes do not stick out or overlap, and all squares are covered by a domino. Merlijn will then color all the dominoes on the board red or blue. Determine the smallest non-negative integer $V$ (depending on $m$ and $n$) such that Merlijn can always ensure that in each row the number of squares covered by a red domino and the number of squares covered by a blue domino differ by at most $V$, no matter how Jetze covers the board.
First, assume that $n$ is odd. Then it is clear that $V \geq 1$; the difference is always odd. We show that $V=1$ is always possible. To do this, color the vertical dominoes in odd columns red and the vertical dominoes in even columns blue. Since in each row, each horizontal domino covers one cell in an even and one cell in an odd column, there is now one more cell covered by a red domino than by a blue domino in each row. Now color the horizontal dominoes in each row alternately blue, red, blue, red, .... If the number of horizontal dominoes is even, there will ultimately be one more red cell; if the number of horizontal dominoes is odd, there will ultimately be one more blue cell. The difference is thus always 1.

If $n \equiv 2 \bmod 4$, then $V \geq 2$ if Jetze places every domino horizontally; then there are an odd number of dominoes in each row. We now show that $V=2$ is always possible. Use the same strategy as in the odd case. After coloring the vertical dominoes, there are an equal number of red and blue cells in each row. If we then color the horizontal dominoes alternately red and blue in each row, the difference between the number of red and blue cells in each row is 0 or 2.

Finally, we consider the case where $n \equiv 0 \bmod 4$. We show that $V=0$ is always possible. Number the rows from top to bottom from 1 to $m$ and let $b_{i}$ be the number of vertical dominoes whose top cell is in row $i$. By induction on $i$, we can easily show that $b_{i}$ is even, using the fact that a horizontal domino always covers an even number of cells in a row. We color the vertical dominoes in row $i$ and row $i+1$ as follows: if $b_{i} \equiv 0 \bmod 4$, we color half red and half blue, and if $b_{i} \equiv 2 \bmod 4$, we color two more red than blue if $i$ is even, and two more blue than red if $i$ is odd. We now show that we can color the horizontal dominoes in each row $k$ so that this row has an equal number of red and blue cells. If $b_{k-1} \equiv b_{k} \equiv 0 \bmod 4$, then the vertical dominoes in row $k$ cover an equal number of red and blue cells. Moreover, the number of horizontal dominoes in row $k$ is now even, so we simply color half red and half blue. If $b_{k-1} \equiv b_{k} \equiv 2 \bmod 4$, then the vertical dominoes in row $k$ again cover an equal number of red and blue cells, because one of $k-1$ and $k$ is even and the other is odd. Furthermore, the number of horizontal dominoes in row $k$ is again even, so we can again color half red and half blue. If $b_{k-1} \not \equiv b_{k} \bmod 4$, the difference between the number of red and blue cells covered by vertical dominoes is 2. The number of horizontal dominoes in row $k$ is now odd. We can therefore color the horizontal dominoes so that ultimately there are an equal number of red and blue cells.

We conclude that the minimal values are: $V=1$ if $n$ is odd, $V=2$ if $n \equiv 2 \bmod 4$, and $V=0$ if $n \equiv 0 \bmod 4$.