We claim that the answer for any $n$ is $n + \bigg\lfloor \frac{n}{2} \bigg\rfloor$ (so in this case it'd be $3031$). This can be achieved by placing a stone in every cell on one of the main diagonals and then placing a stone in every cell on one half of the other main diagonal so that the stones for a lowercase "y" shape. An example is shown for $n=5$ $$\begin{bmatrix} \ast & . & . & . & \ast \\ . & \ast & . & \ast & . \\ . & . & \ast & . & . \\ . & \ast & . & . & . \\ \ast & . & . & . & . \end{bmatrix}$$ This works because if we start by placing a stone on every cell on one of the main diagonals, as shown here $$\begin{bmatrix} . & . & . & . & \ast \\ . & . & . & \ast & . \\ . & . & \ast & . & . \\ . & \ast & . & . & . \\ \ast & . & . & . & . \end{bmatrix}$$ The only pairs of cells with the same set of stones left are those that are symmetric WRT to the diagonal of stones. However, after we add in the stones on the other half diagonal, notice that every cell in the upper-left contains at least one of those stones in the half-diagonal in its row or column. As it already has a stone in its row and column, now every cell in the upper-left has at least two stones in its row or column. But we know that stone symmetric to it is not in the same row or column, so therefore it cannot contain both of these stones, as desired $$\begin{bmatrix} . & . & \color{blue} . & . & \color{red} \ast \\ . & . & . & \ast & . \\ . & . & \color{red} \ast & . & \color{blue} . \\ . & \ast & . & . & . \\ \ast & . & . & . & . \end{bmatrix} \rightarrow \begin{bmatrix} \color {red} \ast & \color{magenta} . & \color{blue} . & \color{magenta} . & \color {red} \ast \\ . & \ast & . & \ast & . \\ . & . & \color {red} \ast & . & \color {blue} . \\ . & \ast & . & . & . \\ \ast & . & . & . & . \end{bmatrix}$$ Now we'll prove this is optimal. Suppose there exists a row without any stones: call this row $A$, and there is also a row with at most one stone: call this row $B$. Then suppose the stone in row $B$ happens to be in column $C$ (if row $B$ is empty, then any column will suffice). Now notice that row $A$, column $C$ and row $B$, column $C$ have the same set of stones (only the stones in column $C$). This means if we have an empty row, every other row must have at least two stones, which leads to at least $2(n-1)$ stones, which is not optimal. Therefore, now we know every that row (and column) has at least one stone in the optimal case. Now call a row or column "deficient" if it only has one stone. Suppose that there are exactly $x$ deficient rows. Then summing through the rows we have $$T \geq x + 2(n-x) = 2n - x \quad (\bigstar)$$where $T$ is the total number of stones. Notice that if row $A$ and row $B$ are deficient rows and both stones are in column $C$, then row $A$, column $C$ and row $B$, column $C$ have the same set of stones (only stones in column $C$) which is an immediate contradiction. Therefore, no two deficient rows can have stones in the same column. Now call a column "special" if it passes through a stone on a deficient row. Since every stone on deficient row is in a different column, there are exactly $x$ special columns. Now suppose two special columns $C$ and $D$ are deficient, and they pass through stones on deficient rows $A$ and $B$ respectively. Finally, row $A$, column $D$ and row $B$, column $C$ have the same set of stones (the two stones on their intersections), which is a contradiction. (In the diagram, rows $A$ and $B$ are the second and fourth rows, and column $C$ and $D$ are the second and fourth column) $$\begin{bmatrix} . & \color {red} . & . & \color {blue} . & . \\ \color {blue} . & \color {magenta} \ast & \color {blue} . & \color{magenta} . & \color {blue} . \\ . & \color {red} . & . & \color {blue} . & . \\ \color {red} . & \color{magenta} . & \color {red} . & \color {magenta} \ast & \color {red} . \\ . & \color {red} . & . & \color {blue} . & . \end{bmatrix}$$ Therefore, only one special column can be deficient. So now summing through the columns we get $$T \geq 1 + 2(x-1) + (n-x) = x + n - 1 \quad (\bigstar\bigstar)$$Finally, adding $(\bigstar)$ and $(\bigstar\bigstar)$ we get $$2T \geq 3n - 1 \Longrightarrow T \geq n + \frac{n-1}{2} \Longrightarrow \boxed {T \geq n + \bigg\lfloor \frac{n}{2} \bigg\rfloor}$$as desired $\blacksquare$