Solution:

Label the rows and columns of the chessboard from $1$ to $8$. The square in row $i$ and column $j$ is black if and only if $i + j$ is even.

Since there is one pawn in each row and column, the pawns occupy the squares $(1, \sigma(1)), (2, \sigma(2)), \ldots, (8, \sigma(8))$ for some permutation $\sigma$ of $\{1,2,\ldots,8\}$.

A pawn is on a black square if $i + \sigma(i)$ is even. Let $B$ be the number of pawns on black squares. Then
$$B = \#\{i : i + \sigma(i) \text{ is even}\}.$$

But $i + \sigma(i)$ is even if and only if $i$ and $\sigma(i)$ have the same parity. That is, $i$ is odd and $\sigma(i)$ is odd, or $i$ is even and $\sigma(i)$ is even.

Let $O$ be the set of odd numbers in $\{1,2,\ldots,8\}$, and $E$ the set of even numbers. There are $4$ odd and $4$ even numbers. Since $\sigma$ is a permutation, $\sigma$ maps $O$ to $4$ numbers, and $E$ to $4$ numbers.

Let $a$ be the number of $i \in O$ with $\sigma(i) \in O$, and $b$ the number of $i \in E$ with $\sigma(i) \in E$. Then $B = a + b$.

But the number of $i \in O$ with $\sigma(i) \in O$ is equal to the number of $j \in O$ with $\sigma^{-1}(j) \in O$, i.e., the number of odd numbers in the image of $O$ under $\sigma$ is the same as the number of odd numbers in the preimage of $O$ under $\sigma$, which is $a$.

Similarly, the number of $i \in E$ with $\sigma(i) \in E$ is $b$.

But $a + b$ is the total number of $i$ such that $i$ and $\sigma(i)$ have the same parity. Since $\sigma$ is a permutation, the number of $i$ with $\sigma(i)$ odd is $4$, and the number with $\sigma(i)$ even is $4$.

Therefore, $a + b$ is even, because $a$ and $b$ are both between $0$ and $4$ inclusive, and their sum is between $0$ and $8$ inclusive, and the only way for $a + b$ to be odd is if one is odd and one is even, but since the total number of $i$ is $8$, and the parity is preserved, $a + b$ must be even.

Alternatively, since the total number of black squares on the board is $32$, and any arrangement of $8$ pawns with one in each row and column must cover an even number of black squares, as shown above.

Thus, the number of pawns on black squares is even.