the following should do:(hopefully i haven't goofed anywhere) consider all pairs of rows. We claim that for some pair of rows, the sum of the entries in those rows is atleast $2n^2$. Indeed, if we denote by $S(i,j)$(should probably denote $S(\{i,j\})$ since the order doesn't matter) the sum of the entries of the $i,j$ rows($i\neq j$), then $\sum_{i\neq j} S(i,j) = (n-1)\sum_{i,j} a_{i,j}=(n-1)n^3$ and since there are $n(n-1)/2$ such terms in the sum, it follows that for some pair $i,j$ the sum of the entries of the corresponding rows is atleast $\frac{(n-1)n^3}{n(n-1)/2}=2n^2$. for this pair of rows, we claim that for some pair of columns $k,l$, the corresponding entries must sum to more than $3n$. Suppose not. again by $T(k,l)$ if we denote the sum of the $4$ entries in the rectangle formed by the rows $i,j$ and columns $k,l$, then we have $T(k,l)\le 3n\ \forall k\neq l$. again consider the sum $\sum_{k\neq l} T(k,l)$;this sum is atleast $(n-1)(2n^2)$(since in this sum, each term of the rows $i,j$ is counted $n-1$ times and so the sum is $(n-1)S(i,j)$). On the other hand, each such term is $\le 3n$ and this counts $n(n-1)/2$ terms, so the total is atmost $3n(n(n-1)/2)$; comparing, we get, $(n-1)(2n^2)\leq 3n^2(n-1)/2 \Rightarrow 2\leq 3/2$ and that is a contradiction.

Unless I'm misunderstanding, in which case my apologies to you, seshadri, I don't think that's it. The four squares with entries which sum up to more than $3n$ must be the vertices of a square, not just a rectangle. In your notation, you should have the additional condition $|i-j|=|k-l|$. A much better bound is easy to prove without this condition: if you sum the sums of all the rectangles (formed by the intersections of two rows and two columns, like your $i,j$ and $k,l$), each unit square is counted $(n-1)^2$ times, so the sum we get is $(n-1)^2n^3$. Since there are $\binom n2^2$ such rectangles, there must be two rows and two columns whose intersections sum up to at least $4n$ (which is pretty intuitive anyway; in fact you proved this stronger bound with your $2\ge\frac 32$, I think ). Again, sorry for any misunderstandings that might have occured.

nope, grobber, i misread about the square part. thanks for pointing it out!

since only squares are to be considered, do the same as i tried before but restricitng only to squares. each cell lies in $\textit{atleast}\ n-1$ squares, so on the one hand the sum over all $2\times 2$ squares is $\ge (n-1)n^3$(this is where positivity of the entries of the array is needed); on the other hand since there are $1^2+2^2+\cdots+(n-1)^2$ squares formed by the cells, the sums are(assuming that the sums are all $\leq 3n$) the total is $\le\frac{(n-1)n(2n-1)}{6}\cdot (3n)$.but then $(n-1)n^3\le\frac{(n-1)n(2n-1)}{6}\cdot (3n)\Rightarrow n\le n-\frac{1}{2}$, a contradiction.