Let $A$ be a symmetric row-column increasing $n \times n$ matrix (i.e. $A(i, j) < A(i+1,j)$ and $A(i, j) < A(i, j+1)$) with integer entries $A(i, j) \in \{1, 2, ..., n^2\}$. Moreover, let us assume that $A$ contains $\Theta(n^2)$ distinct entries, so the set of entries has positive density. 


Define a "row discrepancy" for a rectangle $R=\{A(i,j); a\_1 \leq i \leq a\_2, b\_1 \leq j \leq b\_2 \}$ as $$D(R) = \frac{(a\_2-a\_1)(b\_2-b\_1)}{\max\_{b\_1 \leq j \leq b\_2} (A(a\_2, j)-A(a\_1, j))}.$$ Let us also define the total "row discrepancy" $D\_r$ as a supremum of $D(R)$ over all rectangles $R$. Is it true that $D\_r$ must go to infinity as $n \to \infty$ for any $A$?



Is a "row discrepancy" of symmetric row-column increasing matrices unbounded?
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 $D(R)\le 10$.


