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, so that every domino covers exactly two squares, dominos do not protrude or overlap, and all squares are covered by a domino. Merlin then moves all the dominoe color red or blue on the board. Find the smallest non-negative integer $V$ (in terms of  $m$ and $n$) so that Merlin can always ensure that in each row the number squares covered by a red domino and the number of squares covered by a blue one dominoes are not more than $V$, no matter how Jetze covers the board.
1. We are given an \( m \times n \) board where \( mn \) is even. This means the board has an even number of squares, which can be completely covered by dominoes, each covering exactly two squares.
2. Jetze covers the board with dominoes such that no dominoes protrude or overlap, and all squares are covered.
3. Merlin colors each domino either red or blue.
4. We need to find the smallest non-negative integer \( V \) such that in each row, the number of squares covered by red dominoes and the number of squares covered by blue dominoes are each at most \( V \).

To solve this, consider the following:
- Each row has \( n \) squares.
- Since each domino covers two squares, the number of dominoes covering a row is \( \frac{n}{2} \) if \( n \) is even, or \( \frac{n+1}{2} \) if \( n \) is odd.

Let's analyze the extreme cases:
- If all dominoes in a row are colored the same (either all red or all blue), then one color will cover \( n \) squares, and the other color will cover \( 0 \) squares.
- Therefore, in the worst-case scenario, one color will cover \( n \) squares in a row.

Thus, the smallest non-negative integer \( V \) that ensures the number of squares covered by each color in any row does not exceed \( V \) is \( n \).

Conclusion:
The smallest non-negative integer \( V \) such that in each row, the number of squares covered by red dominoes and the number of squares covered by blue dominoes are each at most \( V \) is \( n \).

The final answer is \( \boxed{n} \)