This problem is cool.

Great problem. Solution from Twitch Solves ISL: Assume that some dominoes are placed so that no more dominoes can be added to the grid. We assume $\min(m,n) \ge 3$ for convenience; the case $m=1$ is straightforward, and the case where $m=2$ can be dealt with by hand. From every unoccupied cell not on the border of the board, we emit four lasers from the center of the cell one in each of the four directions; these must each hit the edge of a domino immediately. Cells on the border emit two or three lasers only, according to whether or not they are on the corner. [asy][asy] size(6cm); for (int i=1; i<6; ++i) { draw((0,i)--(6,i), grey); draw((i,0)--(i,6), grey); } draw(box((0,0),(6,6)), black); path vdomino = box((0.1,0.1), (0.9,1.9)); filldraw(shift(0,1)*vdomino, lightcyan, blue); filldraw(shift(0,4)*vdomino, lightcyan, blue); filldraw(shift(1,0)*vdomino, lightcyan, blue); filldraw(shift(1,3)*vdomino, lightcyan, blue); filldraw(shift(2,1)*vdomino, lightcyan, blue); filldraw(shift(2,4)*vdomino, lightcyan, blue); filldraw(shift(3,0)*vdomino, lightcyan, blue); filldraw(shift(3,3)*vdomino, lightcyan, blue); filldraw(shift(4,1)*vdomino, lightcyan, blue); filldraw(shift(4,4)*vdomino, lightcyan, blue); filldraw(shift(5,0)*vdomino, lightcyan, blue); filldraw(shift(5,3)*vdomino, lightcyan, blue); void draw_laser(pair O, real r) { dot(O+(0.5,0.5), red); draw((O+(0.5,0.5))--(O+(0.5,0.5)+0.55*dir(r)), red, EndArrow); } draw_laser((0,0), 0); draw_laser((0,0), 90); draw_laser((0,3), 0); draw_laser((0,3), 90); draw_laser((0,3), 270); draw_laser((2,0), 0); draw_laser((2,0), 90); draw_laser((2,0), 180); draw_laser((2,3), 0); draw_laser((2,3), 90); draw_laser((2,3), 180); draw_laser((2,3), 270); draw_laser((4,0), 0); draw_laser((4,0), 90); draw_laser((4,0), 180); draw_laser((4,3), 0); draw_laser((4,3), 90); draw_laser((4,3), 180); draw_laser((4,3), 270); draw_laser((1,2), 0); draw_laser((1,2), 90); draw_laser((1,2), 180); draw_laser((1,2), 270); draw_laser((1,5), 0); draw_laser((1,5), 180); draw_laser((1,5), 270); draw_laser((3,2), 0); draw_laser((3,2), 90); draw_laser((3,2), 180); draw_laser((3,2), 270); draw_laser((3,5), 0); draw_laser((3,5), 180); draw_laser((3,5), 270); draw_laser((5,2), 90); draw_laser((5,2), 180); draw_laser((5,2), 270); draw_laser((5,5), 180); draw_laser((5,5), 270); [/asy][/asy] Claim: Every domino in the grid is hit by at most four lasers; this bound improves to three if the domino is on the border of the grid. Proof. Each edge of the domino is hit by at most one laser. $\blacksquare$ Claim: [Laser double counting] The number of dominoes is at least $\frac{mn-1}{3}$. Moreover, for equality to occur, all four corners must be vacant, and the short end of every domino must either touch the edge of the grid or an adjacent square. Proof. We make the additional observation that the number of dominoes on the border is at least the number of vacant squares on the border. Indeed, if one travels around the perimeter of the border, between any two vacant squares there is some domino present. Let $D$ be the number of dominoes, so $mn-2D$ is the number of vacant squares. Let $Y$ be the number of dominoes which touch the border, and $X$ the number of vacant cells on the border, so that $X \le Y$. Finally, let $0 \le e \le 4$ be the number of vacant corners. By double counting the lasers, we now have \[ 4(mn-2D) - X - e \le 4D - Y. \]Since $X \le Y$ and $e \le 4$, we recover $3D \ge mn-1$ as claimed, with equality if $X=Y$, $e=4$, and every edge of every domino is either on the border or touched by a laser. $\blacksquare$ Claim: We can't have $3D = mn-1$ exactly. Proof. We take the standing assumptions from the previous claim that all four corners are empty and every edge of every domino is hit by a laser unless it's at the edge of board. Consider the northwest vacant cell $A$; it is border by two different dominoes. We consider two different possibilities illustrated below for the layout of the two dominoes touching $A$. [asy][asy] picture case1; picture case2; path vdomino = box((0.1,0.1), (0.9,1.9)); size(10cm); for (int i=1; i<6; ++i) { draw(case1, (0,i)--(6,i), grey); draw(case1, (i,0)--(i,6), grey); } draw(case1, (0,0)--(0,6)--(6,6), black); label(case1, "$A$", (0.5,5.5), red); filldraw(case1, shift(0,3)*vdomino, lightcyan, blue); case2 = rotate(0)*case1; filldraw(case1, shift(1,4)*vdomino, lightcyan, blue); label(case1, "$1$", (2.5,5.5), red); label(case1, "$2$", (1.5,3.5), red); label(case1, "$3$", (0.5,2.5), red); label(case1, "$4$", (2.5,2.5), red); filldraw(case1, shift(2,3)*vdomino, palegreen, deepgreen+dashed); filldraw(case1, shift(1,1)*vdomino, palegreen, deepgreen+dashed); filldraw(case2, shift(1,5)*reflect( (0,0), (1,1) )*vdomino, lightcyan, blue); label(case2, "$5$", (1.5,4.5), red); filldraw(case2, shift(1,2)*vdomino, palegreen, deepgreen+dashed); filldraw(case2, shift(2,4)*reflect( (0,0), (1,1) )*vdomino, palegreen, deepgreen+dashed); label(case2, "$6$", (2.5,3.5), red); label(case1, "First case", (3,0), dir(-90)); label(case2, "Second case", (3,0), dir(-90)); add(shift(-8,0)*case1); add(shift(0,0)*case2); [/asy][/asy] In first case, the dominoes point the same way, say vertically. Consider the cell marked $1$; it must be vacant (since either a horizontal or vertical domino would violate the equality case condition). The cell immediately below $1$ must be occupied, and again we see the domino should be vertical. This means the cells marked $2$, $3$, $4$ indicated must be vacant, and we get a vertical domino touching all three. This then lets us repeat the process to find that all dominoes in the first three columns are vertical. We can then extend the process to the right as well; eventually, we find all dominoes are vertical. But then $D > mn/3$ can clearly not hold. The second case is easier: the cell marked $5$ must be vacant, and we must have the two dominoes shown since the short ends of these dominoes cannot touch the blue ones. Then the cells marked $6$ is vacant, and we can repeat this argument. Eventually though we reach the edge of the board and a contradiction takes place. $\blacksquare$

This turns out to have an OEIS sequence.

Assume $m$ and $n$ are even for now. Define the $A$ and $B$ partitions to be the two partitions in the attached picture. A domino is called "type $a$" if it does not intersect any of the lines in partition $A$ and "type $b$" if it does not intersect any of the lines in partition $B$. Every possible domino on the board is one of the two types. Suppose there are $k$ dominoes on the board and no more dominoes can be added. Assume wlog that the number of dominoes $d$ of type $a$ is at least the number of dominoes of type $b$. To say, $d \geq \frac{k}{2}$. Now, notice that we can completely tile with $m \times n$ board with phantom dominoes where no two phantom dominoes overlap and where each original domino of type $a$ coincides with a phantom domino. indeed, this is achieved by tiling each $2 \times 2$ square vertically or horizontally, which is parallel to the type $a$ dominoes in the $2 \times 2$ square. Now, each domino of type $b$ intersects at most $2$ phantom dominoes. Notice that if there is some phantom domino that is not intersected by any of the original dominoes, then we can place a new domino where that phantom domino is, contradicting the fact that we could not place any new dominoes. Since there are $d$ dominoes of type $a$, $k-d$ dominoes of type $b$, and $\frac{mn}{2}$ phantom dominoes, we get the following inequality \[ \frac{mn}{2} \leq d + 2(k-d) = 2k-d \leq 2k-\frac{k}{2} = \frac{3}{2}k \]So, we have $k \geq \frac{mn}{3}$ as desired. A very similar sort of argument works for $m$ or $n$ not even, but dealing with the $1 \times 1$ gaps take some care. Remark: I know I wrote partion instead of partition in my diagram. We can all ignore this.

Excellent problem! Fill in each empty cell with a square piece of cheese. Draw a bipartite graph between cheese pieces and dominoes, drawing an edge if a cheese and domino are adjacent. Suppose that there are $A$ cheeses and $B$ dominoes; our goal is to prove that $A \leq B$. Every interior, edge, or corner cheese must have degree $4,3,2$ respectively, hence there are $4A-E-2C$ edges total where $E,C$ count the number of edge and corner cheeses. Furthermore, by similarly considering domino positions, there are at most $4B-D$ edges total, where $D$ is the number of domino edges that lie on the border. Suppose that some border edge touches $k$ non-corner cheeses. Then since there is at least one edge domino (possibly a corner domino) between any two non-corner cheeses on this edge, as well as an edge domino each on the left and right of the leftmost and rightmost edge cheeses. Across all edges, each domino gets counted at most as many times as it has border-edge touches (for example, a corner domino would have be counted at most twice). Hence by looking at dominoes, there are at most $4B-E-4$ edges, so $$4A-E-8 \leq 4A-E-2C \leq 4B-E-4 \implies A \leq B+1.$$Therefore, it suffices to prove that $A=B+1$ is impossible. Note that equality holds if and only if each domino has number of edges exactly equal to $4$ minus the number of border-edge overlaps. In particular, the "short edge" of every domino must either touch the border or some cheese. Suppose that such a configuration existed. Suppose some cheese square $c$ has two differently-oriented dominoes sharing a short edge with it (forming a sort of $L$ shape). Consider the cell $c'$ that's the reflection of $c$ over the shared corner of these two dominoes, which must be cheese. Due to equality we can see that the other two edges of $c'$ border the short sides of dominoes. Repeating this, we eventually arrive at a contradiction when we run into the edge of the grid. Using the above result as well as the initial "short edge" condition, it is not hard to see that every non-corner cheese must have exactly two similarly-oriented dominoes sharing a short edge with it. Starting from an arbitrary corner and repeating this implies that every domino has the same orientation (WLOG vertical). If $m$ is the number of rows, then by looking at the leftmost row the requirement that we have four corner cheeses implies that $m \equiv 1 \pmod{3}$. But then we reach a contradiction on the second-leftmost row, since its topmost cell is part of a vertical domino and thus we need $m \equiv 2,3 \pmod{3}$. Thus equality cannot occur, so $A \leq B$ as desired. $\blacksquare$

Call an empty cell a void. Note that each void is bordered by $4$ dominos. Call a chain of alternating voids and dominos a zigzag, where an adjacent domino / chain lies above or below a domino. Note that the condition implies that if a zizag ends with a void, that void must be on the edge of the chessboard. Take zigzags of maximal length. Define the difference of a zigzag as the number of dominos minus the number of voids. Note that the difference of a zigzag is then $1 - e$ where $e \in \{0, 1, 2\}$ is the number of ends of the zigzag which are voids. Now, suppose that there are $T$ voids on the top of the boards and $R$ voids on the bottom of the board. Let there be $d_T$ and $d_R$ on the top and bottom too. Note that $d_T \ge T - 1, d_R \ge R - 1$ There must be at least $\max\{2|R| - 1, 2|T| - 1\}$ zigzags, so the sum of differences over all zigzags is $\max\{R + d_R, T + d_T\} - |R| - |T|$. This is always positive except for if $R = T, d_R = d_T = R - 1$. Then the $2R - 1$ zigzags from the top dominos/voids to the bottom must also be the only zigzags, and the top/bottom dominos alternate exactly. The same applies after rotating the board $90$ degrees. Claim: Suppose a domino is horizontal and the two squares above it are empty. Then one of the squares above it is a void and the other is a horizontal domino. Proof. The domino can't be the end of a zigzag beacuse it isn't touching a border. As such, one of the two squares must be a void to continue the zigzag. The other is part of some domino. This domino can't be vertical because then it would start a new zigzag, so it must be horizontal. $\blacksquare$ WLOG rotate such that some domino is vertical. As such, given a vertical domino, we get a horizontal stack of vertical dominos on top of each other from top to bottom, offset by at most $1$. If there were any horizontal dominos, this would lead to a contradiction because then the two stacks would intersect. As such, all dominos are vertical. However. However, this contradicts the fact that there exists two zigzags of the same length, one of which both starts and end with a domino, and the other starts and end with a void.