$25$ is the maximum possible and can be proven by case distinction.
First, notice that by a symmetry argument, we can reduce the number of placements for rook to $10$. Here are the possible placements for rook:
Now, there are few observations to make and few facts to be mentioned:
Let $K(n)$ be the maximum number of non-attacking knights that can be placed to $n\times n$ board. Then (source),
$$K(n) =
\begin{cases}
\frac{n^2}{2}, & \text{if $n$ is even} \\[2ex]
\frac{n^2+1}{2}, & \text{if $n$ is odd}
\end{cases}$$
When we divide a board into $m$ rectangular boards with $K_i$ being the maximum number of non-attacking knights that can be placed to $i^{th}$ small board, we have
$$K \le \sum_{i = 1}^m K_m$$
where $K$ is the maximum number of non-attacking knights that can be placed into non-divided board.
We might have different upper bounds for different board divisions. But in the end, we are trying to find the least upper bound.
When we have a $4\times n$ board with $n \ge 2$, maximum number of non-attacking knights that can be placed $K_4(n) = 2n$. For even $n$'s, this is really easy to prove by dividing the board into $2\times4$ small boards. Then we already know $K_4(2) = 4$ (my previous answer). For odd $n$'s, first we should observe $K_4(3) = 6$ (can be seen by trial and error). Then for $K_4(5)$, we are just adding a $2\times4$ board piece to $3\times4$ board. Here, you can use induction to obtain the result.
When we have a $1 \times n$ board, obviously, maximum number of non-attacking knights that can be placed is $n$. But we will try to avoid divisions where we have this case. Instead, we will try the next one.
When we have have a case as in the below where we can't place knights to red line, we can consider left side and right side of the line as a single $2 \times k$ board piece. This is because when we have a $2\times k$ board, whenever we place a knight to left $1\times k$ part, it threatens $1$ or $2$ squares in right $1\times k$ part. Notice that this is the same for the below case.
And in order to find maximum number of non-attacking knights that can be placed, we can divide the board (green lines divides it) as in the figure and see that for each $4$ squares between the green lines, we can place $2$ knights.
When we have $3 \times 5$ board, we can place at most $8$ knights. This is also worth mentioning because we will see this case a lot. In order to prove, we can divide it into $3 \times 3$ where maximum is $5$, and $2 \times 3$ where maximum is $4$. So our upper bound is $9$. But, there is only two ways of placing $5$ knights to $3 \times 3$ board (shown below). In both cases, you can see that we can't place more than $3$ knights to $2 \times 3$ board. Therefore, $9$ is not achievable. Note that this will be our main proof idea from here onward.
Now, we can look at the cases. I will write the maximum number of knights to each division and red notches will indicate where knights can threaten the rook (therefore we can't place knights):
Case 1: Rook in a1. This is an easy case since it leaves us with $7 \times 7$ board where we can't place knights to two of white squares. but on that $7 \times 7$ board, there are $25$ black squares. So we have $K_{a1} \le K(7) = \frac{7^2+1}{2} = 25$ by the first fact and indeed we can place $25$ knights as you suggested.
Case 2: Rook in b1. We can divide the board as in the following:
Here, we have $26$ as upper bound. However, this is still not bad because in order to achieve $26$, all the divisions must have maximum possible of knights. Therefore, whenever we prove we can't achieve maximum for a division, upper bound becocmes $25$.
Now, for blue part of the board, placement of $8$ knights is unique (shown below). And when we make that placement, $3$ squares of $3 \times 5$ board is threatened (shown by black notches). And dividing it into a $3 \times 3$ and a $2 \times 3$ board, if we use all $3$ unthreatened squares of $2 \times 3$, we can't place $5$ knights to $3 \times 3$ board. Therefore, we can't place $8$ knights to blue part and $3 \times 5$ board. Therefore, $26$ is not achievable in this case.
Case 3: Rook in b2. We can divide the board as in the following:
Here, we have $K_{b2} \le 25 = 13+6+4+2$.
Case 4: Rook in c1. We can divide the board as in the following:
Here, again we have $26$ as an upper bound. But if we focus left side of the board, we can see that if we place $2$ knights to $2 \times 2$ board, we can't place $6$ knights to $2 \times 5$ board (shown below). Therefore, $26$ is not achievable in this case.
Case 5: Rook in c2. We can divide the board as in the following:
Here, we have $26$ as an upper bound. But green part has a unique placement for $6$ knights (shown below). After that placement, we can observe that for $2 \times 5$ board, placement of $5$ knights is determined. After that, we should notice we can't place $8$ knights to $3 \times 5$ board. Therefore, $26$ is not achievable.
Case 6: Rook in c3. We can divide the board as the following:
Then, we have $K_{c3} \le 25 = 13+5+5+2$.
Case 7: Rook in d1. We can divide the board as the following:
Then, we have $K_{d1} \le 25 = 14+6+5$.
Case 8: Rook in d2. We can divide the board as the following:
Here, again we have $26$ as an upper bound. But it is easy to see that if we place $4$ knights to blue part, we can't place maximum of $2$ knights to $2 \times 1$ board. Therefore, $26$ is not achievable in this case.
Case 9: Rook in d3. We can divide the board as the following:
Then, we have $K_{d3} \le 25 = 10+8+4+3$.
Case 10: Rook in d4. We can divide the board as the following:
Then, we have $K_{d4} \le 25 = 8+6+6+5$.
Therefore, the maximum possible number is $25$.
Unfortunately, this is a very long solution and in some of $10$ cases, we found upper bound to be $26$ and forced to prove $26$ was not possible (maybe there are some divisions of board where $25$ is upper bound for these bad cases). I couldn't see a way to show $25$ is the maximum possible without a case distinction of rook placements. If there is an easier way, I am also glad to see that.
Best Answer
Note that when we have two knights threatening each other, it actually forms either a $2\times3$ or $3 \times 2$ board. And for each of $2 \times 3$ and $3 \times 2$ boards, there are $2$ ways of placing two knights so that they threaten each other. So, what we should do is to count how many $2 \times 3$ and $3 \times 2$ squares on $n\times n$ board. For general $n$, the answer is $$(n-1)(n-2)+(n-2)(n-1) = 2(n-1)(n-2)$$ And for each $2\times3$ and $3\times2$ board, there are $2$ ways of placing the knights so that they threaten each other. Therefore, in total there are $$2\cdot2(n-1)(n-2)=4(n-1)(n-2)$$ ways of placing two knights so that they threaten each other. So what you are looking for is $$\frac{n^2(n^2-1)}{2}-4(n-1)(n-2)$$ It is also worth mentioning that we are not over-counting because whenever we place two knights so that they threaten each other, either a $2 \times 3$ or $3 \times 2$ board must contain both of the knights.