Let $A\in M_n(\Bbb R)$ be such that the sum of the two largest numbers in each row is $a$, and in each column is $b$. How can I prove that $a=b$? 
In every cell of a square table there is a real number. The sum of the largest two numbers in each row is $a$ and the sum of the largest two numbers in each column is $b$. Prove that $a = b$.

Help! I don't know how to start the problem. I've considered the largest and smallest numbers of such a set, however, I don't know what good using the extremes will bring.
 A: HINT: Suppose that $a>b$. Let $d=\frac14(a+b)$, and subtract $d$ from each cell of the table. On the one hand $d<\frac{a}2$, so the sum of the largest two elements in each row is positive, but on the other hand $d>\frac{b}2$, so the sum of the largest two elements in each column is negative.
Added: I initially thought that it was almost done at this point, but as Calvin Lin pointed out, we cannot conclude that every column consists entirely of negative numbers and get an immediate contradiction. However, we can immediately conclude that each row contains at least one positive number, and each column contains at most one positive number. This implies that there is exactly one positive number in each row and column.
Now let $x$ be the largest negative number in the table (i.e., the one with the smallest absolute value); say that $x$ is in column $c$. Let $y$ be the positive number in column $c$, say in row $r$. Finally, let $z$ be the largest negative number in row $r$. Then $z<x$, so $0<z+y<x+y<0$, which is absurd. Thus, $a\le b$, and by symmetry $b\le a$, so $a=b$.
A: For each row $i$, let $x_i = x_{i,j(i)}$ be the largest element and $x'_i = x_{i, j'(i)}$ be the second largest element, breaking ties arbitrarily if necessary.
Hint:  Show that $ b \geq a$.
Proof:  WLOG, $x_1$ is the smallest of the $x_i$.
Case 1: There is a column with 2 of these elements $x_i$, say $x_k$ and $x_l$:

 then $b \geq x_k + x_l \geq x_1 + x_1 \geq a  $.

Case 2: If not, then each $x_i$ is in it's own column.

 Consider $ x'_1 = x_{i, j'(1)}$.
 Consider column $j'(1)$, which has corresponding circled number $x_{j'(1)} $.
 Then,  $b \geq x_{j'(1)} + x'_1 \geq x_1 + x'_1 = a $

Corollary: By symmetry, $ a \geq b$ hence $ a = b$.
