4. Fill the positive integers $1,2,3,4,5,6,7,8,9$ in a $3 \times 3$ grid, using each positive integer exactly once, such that the sum of the three numbers in each row and the sum of the three numbers in each column are all odd. The total number of such arrangements is $\qquad$ kinds.
The number of odd numbers in each row and each column is an odd number. With 5 odd numbers, the distribution for each row and each column is $1,1,3$. Therefore, the 5 odd numbers can only be placed in 5 cells of a $3 \times 3$ grid, one row and one column.

Thus, the total number of ways to satisfy the condition is $C_{3}^{1} \cdot C_{3}^{1} \cdot 5! \cdot 4! = 25920$.