Note that the only way to get an odd sum is if there are either $0$ or $2$ even numbers in the row or column.

The only way for this to happen is if the $4$ even numbers form a rectangle with sides parallel to the large square.

The way to see this is we choose a spot for the first even number. Then we need to choose another square in the same row, $x,$ and column, $y,$ to be even.

The final even has to be in same column as $x$ and the same row as $y.$ This forms the aforementioned rectangle.

There are four $2 \times 2$ rectangles, two $3 \times 2$ rectangles, two $2 \times 3$ rectangles, and one $3 \times 3$ rectangle.

This gives a total of $$4+2+2+1=9$$ possible sets of positions for the even numbers.

There are $4!$ ways to arrange the even numbers and $5!$ ways to arrange the odd numbers.

This means that there are a total of $$9 \cdot 4! \cdot 5!$$ configurations of squares that satisfy the condition.

There are a total of $9!$ arrangements with no restrictions. The probability is therefore $$\dfrac{9 \cdot 4! \cdot 5!}{9!} = \dfrac{1}{14}.$$

Thus, B is the correct answer.