I claim the answer for odd $n$ is $\frac{3n-1}{2}$ and for even $n$ is $\frac{3n-2}{2}$. I will solve only for odd $n$ since the solution for even is very similar. This can be attained by putting $n$ rooks in the main diagonal and $\frac{n-1}{2}$ in the first half of the other diagonal. Now, suppose we can put more than $\frac{3n-1}{2}$ rooks in the table. Since no row or column can have more than $2$ rooks, then there are going to be at least $\frac{n+1}{2}$ rows with $2$ rooks. Now we introduce the following lemma: Lemma: There cannot be more than $\frac{n-1}{2}$ rows with more than $1$ rook. Proof: Suppose the rooks in those rows are ordened by the distance $d_i$ between two of them. Note these distances all, if ordened in a list, show us that $d_{i-1}-d_i \geq 2$, as a simple diagram can show two rooks would not attack a common square. The result follows.