Does anyone have a solution?

I have the following argument but am not certain of correctness or rigor.

Another way to show you can't visit both the upper left and the bottom right corner is by considering leopard cycle as a non-intersecting polygon $\mathcal{P}$ with arrows on the sides representing the direction the leopard was moving in. Paint the interior of $\mathcal{P}$ red (the rest of the board is white say). Consider each of the arrows drawn on the side of $\mathcal{P}$. The red region will always lie on the same side with respect to the arrow (right or left). If both the upper left and the bottom right corner are visited by the leopard, then by the previous argument there must be a region outside of the chequer board painted red, contradiction. This is because the only way to visit these cells is using the simple moves defined above.

Solved with Jeffrey Chen The answer is $9n^2 - 3$. The construction is as follows: Call the operation of going right then diagonal as a "rd" operation, while going up then diagonal a "ud" operation. Start at the top left, then go one space to the right. Now, perform $3n - 3$ rd operations, go forward twice, go up $3n - 2$ times, then go forward again. Repeat this until you hit the right wall, and can not move forward. Then, go diagonal twice, and now perform $3n-3$ ud operations. Finally, keep going up until you reach our starting point. This clearly hits $9n^2 - 3$ squares. Now, I claim this is maximal. Since we go back to where we started, the number of up, right, and diagonal moves are the same, so the square we visited is divisible by $3$. FTSOC assume such a cycle existed. Now, if we visited more than $9n^2 - 3$ squares, we must've visited every square. WLOG, let us go into the top right square using a right operation. Then, the only way to exit is to use a diagonal operation. Next, If we do not move right, then the square to the right of our current square, it will have to exit with a diagonal. Now, two vertically adjacent squares were entered by a diagonal, so they must again exit with a diagonal. This continues until we hit the bottom side, so the cycle can't be completed, a contradiction. Thus, we must move right, and then move diagonal again, forming an rd operation. By the same logic, we will continue to use rd operations until we hit the bottom right square, at which point we can never exit, a contradiction. Therefore, there are no cycles that hit every cell, so the maximal moves the leopard could make is $9n^2 - 3$.

Solved with cursed_tangent1434. This problem played with us for a while until we actually realized what was going on. Regardless it's still really beautiful. We claim that the answer is $M(3n)=9n^2-3$ squares. Throughout this solution, we denote by $(i,j)$ the square which is at the intersection of the $i^{\text{th}}$ row and $j^{\text{th}}$ column. We first show that this answer is indeed achievable. The Construction: We first show that it is possible to start from cell $(1,1)$ visit every single cell and finish at cell $(1,3m)$ for any $n \times 3m$ rectangle. To do this, first consider a $n\times 3$ rectangle. We do the following 'zig-zag' moves. \[(1,1) \to (1,2) \to (2,1) \to (2,2) \to \dots \to (n,1) \to (n,2)\]Then, we move to the square $(n,3)$ which is to the right of this square and move along the $3^{rd}$ column right to the top until we reach square $(1,3)$ as desired. Now, it is easy to see that this extends to arbitrary $n \times 3m$ rectangles since we simply have to move from the top-right corner of one $n \times 3$ to the top-left corner of the adjacent one. Now, we start from square $(1,3)$ and move diagonally through squares $(2,2)$ and $(1,3)$. Then, using the previous result, we can visit every cell of the bottom $3n-2 \times 3n$ rectangle and reach $(3,3n)$. Then, after moving to $(2,3n)$ we can again apply the result on the top-right corner $2\times 3n-3$ rectangle and reach square $(1,4)$ we then move to $(2,3)$ and $(1,3)$ returning to our original position, covering all squares except the three at the top-left hand corner. Proof of Bound: Next, we show that this is optimal. We define a $3-$chess coloring of the $3n\times 3n$ board, where the board is colored in 3 colored red ($R$) , green ($G$) and blue ($B$). Here, we color the first column, in the pattern $R-G-B$ downwards. Then, for every row starting with $R$ we color in the order $R-B-G$ , for every row starting with $G$ we color $G-R-B$ and for every row starting with $B$ we color $B-G-R$. Then, it is not hard to see that a leopard which is on a $R$ square must move to a $B$ square, a leopard on a $B$ square to a $G$ square and a leopard on a $G$ square to a $R$ square. Now, since the current square color is periodic $\pmod{3}$, it follows that if we start from a certain square and return to it (no nte change in color) then we must have moved a number of moves which is $\pmod{3}$ and since we don't visit the same square twice, we must have visited a number of squares which is $\pmod{3}$. Thus, if we can do better than our claimed answer, we must be able to achieve $M(3n)=9n^2$, i.e there exists a cycle which passes through every single cell on the grid. Now suppose FTSOC such cycle covering all the grid exists, notice that only way to exit $(3n,3n)$ is down-left and the only two ways of entering it are up or right. Now we will draw an arrow point out the direction in which the leopard moved on the cycle and we will now give two claims that come from the assumption. Claim 1: In the cycle, after entering $(1,3n)$ the only way to enter $(3n,3n)$ is through the right. Proof: Suppose FTSOC we enter using the move "up", in that case consider the path the leopard took to arrive the cell below $(3n,3n)$. Consider the upper-most right arrow drawn in each column as we as the downcleft ones and the up ones. Since we are given a cycle we should be able to start anywhere and still come back so we will start at $(1,3n-1)$ in this case (next moves are forcefully $(1,3n)$ and then $(2,3n)$, and the only way to enter $(1.3n-1)$ is $(1,3n-2)$). Now lets draw a "closure" which consists in painting all the squares that were used in the path before entering $(3n,3n)$. Note from $(3n,3n)$ the only way to go back to $(1,3n-1)$ is by a sequence of left-down's and other things, but we can never entered the "closure" as the leopard cannot skip any "left-down" diagonal on its path due to its limited movement and grids with "up" or right" also block any way to enter "diagonally" which is always forced at some point or else you never get to column $1$. Finally because after 2 moves you enter $(2,3n)$ going diagonal down makes a cycle so next move is forcefully $(3,3n)$ and two diagonal left's can only go to at most $(1,3n-2)$ so that only entrance square is blocked and cannot be entered again from $(3n,3n)$ contradiction!. Claim 2: In the cycle after entering $(3n,1)$ we can only enter $(3n,3n)$ using as last move "up". Proof: Suppose FTSOC you enter through the right, now notice that the proof of this claim is literaly just applying symetry + 90 degrees clock-wise rotation of Claim 1: (notice this swaps the movement like "right" to "up", "up" to "right" and "down-left" is fixed after this transform) (i.e. make a new "closure", focus on the rightmost "up" arrows in each row) this claim is basically analogous and since its a cycle we can start from $(3n-1,1)$ and we should be able to to re-enter it back, but just like in Claim 1 we can't if last move to enter $(3n,3n)$ is "right" due to the new closure, so we enter $(3n, 3n)$ through the last move before it being "up". The finish: Notice that both Claim's contradict each other, because if from $(1,3n-1)$ you entered $(3n,3n)$ but not $(3n,1)$ yet, then $(3n,1)$ is blocked by the sequence of "up"'s in column $3n$ so you must've entered $(1,3n-1)$ before, but now Claim 2 says you enter $(3n,3n)$ from the last move being "up" which contradicts Claim 1 which says we should enter $(3n,3n)$ from "right", so at some point of the cycle you've entered $(3n,3n)$ twice, contradiction!. Therefore $9n^2$ is not achievable and thus $M(3n)=9n^2-3$ squares is maximal thus we are done .