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 $a=b$?

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$.