2019 AMC 10A Problem 20

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.