7. Fill the 16 positive integers from $1 \sim 16$ randomly into the 16 squares of a $4 \times 4$ chessboard (one number per square). The probability that the sum of the numbers in each row and each column is even is $\qquad$
7. $\frac{41}{2145}$.

First, color a $4 \times 4$ chessboard with black and white such that there are 8 black and 8 white squares, and each row (or column) contains an even number of squares of the same color.
We consider three cases:
(1) If the first column has two black and two white squares, then there are $\mathrm{C}_{4}^{2}$ ways to color this column.
Considering the number of black squares in each row of the remaining three columns, there are
$$
1+2 \times 3 \times 2+3 \times 2 \times 2+3 \times 3=34
$$

ways to color them.
(2) If the first column has four black squares, then the remaining three columns have
$$
3+\mathrm{C}_{3}^{2} \mathrm{C}_{4}^{2}=21
$$

ways to color them.
(3) If the first column has four white squares, then the remaining three columns have 21 ways to color them.

For each of these coloring methods, fill the black squares with the even numbers from $1 \sim 16$ and the white squares with the odd numbers, resulting in a valid arrangement.
Thus, the required probability is
$$
\frac{(6 \times 34+21 \times 2) \times(8!)^{2}}{16!}=\frac{41}{2145} .
$$