Contest math problem about crossing out numbers in the table

Let $a_1,a_2\dots a_n$ be the numbers selected by Ann and $b_1,b_2,\dots b_n$ be the numbers selected by Benjamin.

Lemma: $a_i\geq b_{n+1-i}$ for all $1\leq i \leq n$.

Proof: Consider the set $A$ of all numbers that were eligible when we selected $a_i$ and the set $B$ of all the numbers eligible when we selected $b_{n+1-i}$. Notice that these two sets intersect (because $n-i+1$ rows are eligible in $A$ and $i$ rows are eligible in $B$, and the same happens with the columns) so the claim is true, since $a_i=\max(A)$ and $b_i=\min B$.

The problem follows from the lemma:

$\sum\limits_{i=1}^n a_i \geq \sum\limits_{i=1}^n b_{n+1-i} = \sum\limits_{i=1}^n b_i$

And in fact to add to @Jorge's answer strict inequality is not always possible: For any integer $n$ there are $n \times n$ tables $A$ that will force Player A and Player B to have the same sum: Let $A$ be any $n \times n$ table where the entries get larger as you head down a column, and to the right on a row: e.g., $A =[a_{ij}]$ where the $ij$-th entry is $i(n+1) + j$. [Note that all entries of $A$ are distinct.] Then both Player A and Player B will end up picking the diagonal elements of $A$: Player A will end up picking the diagonals southeast to northwest when Player B northwest to southeast.