The answer is $1011^4$. In general, for $m$ rows and $n$ columns, the answer is $$\lfloor \frac{m^2}{4}\rfloor \lfloor \frac{n^2}{4}\rfloor$$ Construction: let $k=\lfloor \frac m2 \rfloor$ and $l=\lfloor \frac n2\rfloor$. Color $(x,y)$ gold iff both or none of $x\le k$ and $y\le l$ holds. We can check after this many moves, we can reach a position where $(x,y)$ is gold iff both or none of $x\le m-k$ and $y\le n-l$ holds. Proof of optimality: let $$S_j=\sum\limits_{(i,j) \text{ black}} i$$Be the "inversion index" for row $j$. Note the sum of $S_j$ is an invariant. We can replace $S_j\rightarrow S_j-c_j$ such that $S_j$ is an integer between 0 and $X$ where $X=\lfloor \frac{m^2}{4}\rfloor$ Consider an alternative process where each move, we increment $S_j$ by $c$ and decrement $S_t$ by $c$ for some $j<t, c\ge 1$ and I get $(t-j)c$ points. It's clear each move gives me at least one point, so it suffices to show that number of points I can earn is at most $X \lfloor \frac{n^2}{4}\rfloor$ WLOG I always decrement $S_j$ and increment $S_{j+1}$. We can see for each $1\le j\le l$ I can score at most $jX$ points by decrementing $S_j$ and incrementing $S_{j+1}$ because it can lose $(j-1)X$ total index from lower rows and $X$ points from this row. Similarly I can score at most $(n-j)X$ points by incrementing $S_j$ and decrementing $S_{j+1}$. Summing, my score (number of points) is at most $X(1+2+\cdots+l+l+\cdots+1)$ for $n$ odd and $X(1+2+\cdots+l+\cdots+1)$ for $n$ even, yields the desired result.