I think all \(mn \equiv 1 \mod 2\) with \(n \geq 3\) work. Or are there more possible values for \(n\)? All I know is that for \(m\) even we can make Breaker pair off the columns and place his blocks on the same height as Maker's in the same pair he put his in. For \(mn\) odd, Maker can make the third row green (just make him put his block on top of Breaker's if Breaker puts a block above the bottom row, and make him put a block in the bottom row otherwise).

Well, $m=1$, $n$ arbitrary is another (trivial) solution, but otherwise that's correct.

Oh, shucks, I forgot about that. And here I was thinking about how I wouldn't miss the trivial case this time Anyways, I think the algorithm for Breaker goes as follows when \(n\) is even and \(m>1\) is odd: Breaker pairs up the 2nd and third columns, fourth and fifth columns, sixth and seventh columns, and so on. If Maker places a block in one of these columns, then Breaker places his in its pair, and if Maker places a block in the first column then so does Breaker.

I think it's better to replace Maker by "Trump Supporter" and Breaker by "Dreamer". Anyway, as the finish from ManuelKahayon strategy for Breaker, here's the strategy for Maker in the case that $m,n\geq 3$ are odd number: Denote the coordinate of each block of the wall by $(x,y)$ where $x\in \{ 1,2,...,m\}$ and $y\in \{ 1,2,...,n\}$. First, he put one green block at $(1,1)$. After that, Maker keeps in mind the following strategic pairings: $$(2i,1)\Leftrightarrow (2i+1,1), (r,s)\Rightarrow (r,s+1)$$for all $i\in \{ 1,2,...,\frac{m-1}{2}\}$ and $r\in \{ 1,2,...,m\} ,n\in \{ 2,3,...,n\}$. Whenever Breaker places red block at one of blocks in the pairs, Maker can places green block at the other. Note that this guarantee that all blocks with coordinates $(x,y)$ with odd $y>1$ will eventually occupied by green blocks (if the game hasn't stop before). So Maker can create a row $y=3$ containing only green blocks, done.