20. Place the numbers $1,2, \cdots, 9$ arbitrarily into a $3 \times 3$ grid, with one number per cell, and each number used only once. The probability that the sum of the numbers in each row and each column is odd is ( ).
(A) $\frac{1}{21}$
(B) $\frac{1}{14}$
(C) $\frac{5}{63}$
(D) $\frac{2}{21}$
(E) $\frac{1}{7}$
20. B.

Notice that, the sum of three numbers is odd if and only if these three numbers are either all odd or two even and one odd.

Thus, the distribution where the sum of numbers in each row and each column is odd is as shown in Figure 5 ("V" marks the even numbers).

Figure 5 Therefore, the required probability is $\frac{4!\times 5!\times 9}{9!}=\frac{1}{14}$.