Put $m:=\lfloor\frac{4n+6}5\rfloor$. We can choose an anti-diagonal strip $S=\{(i,j)\in D\;|\;m \le i+j<2m\}$. Then $S$ is obviously "sum-free", and $|S|=n^2-\binom {m-1}2-\binom {2n-2m+2}2$, which turns out to be equal to $\lfloor\frac{3n(n+1)}{5}\rfloor $ for all $n$ (looking $\mod 5$). It seems like this bound is best possible, although, at least for $n\ge6, n\equiv 1,2\pmod 5$, the set $S$ is not unique.