Maybe I misunderstood something, but if $A(i,j)=in/10+j$, then $A(a_2,j)-A(a_1,j)=(a_2-a_1)n/10$ (plus/minus 1, if you want integers and round the $A(i,j)$'s) and since $b_2-b_1\le n$, we get that $D(R)\le 10$.

This is my updated answer, probably still wrong somewhere...

If $A(i,j)=\lfloor (i+j)n/10\rfloor $, then the matrix is symmetric and satisfies strict inequalities if $n$ is at least $10$.

Also $A(a_2,j)-A(a_1,j)=(a_2-a_1)n/10$ (plus/minus 1, if you want integers and round the $A(i,j)$'s) and since $b_2-b_1\le n$, we get that $D(R)\le 10$.