To prove that the sum of the largest two numbers in each row is equal to the sum of the largest two numbers in each column, we will use induction on $n$, the size of the $n \times n$ square table.

### Base Case: $n = 2$
For a $2 \times 2$ table, let the entries be:
$$\begin{bmatrix} a_{1,1} & a_{1,2} \\ a_{2,1} & a_{2,2} \end{bmatrix}$$

The sum of the largest two numbers in each row is:
$$a = \max(a_{1,1}, a_{1,2}) + \min(a_{1,1}, a_{1,2}) + \max(a_{2,1}, a_{2,2}) + \min(a_{2,1}, a_{2,2})$$

The sum of the largest two numbers in each column is:
$$b = \max(a_{1,1}, a_{2,1}) + \min(a_{1,1}, a_{2,1}) + \max(a_{1,2}, a_{2,2}) + \min(a_{1,2}, a_{2,2})$$

Since the sum of all entries in the table is the same for both rows and columns, we have:
$$a = b$$

### Inductive Step
Assume the statement holds for an $n \times n$ table. We need to show it holds for an $(n+1) \times (n+1)$ table.

Let $X$ be the largest entry in the $(n+1) \times (n+1)$ table. Without loss of generality, assume $X$ is in the bottom left corner. We can shift rows and columns to place $X$ at position $(n+1, 1)$.

Let $C$ be the assertion that there exists another column $c \neq 1$ with one of the largest two numbers in the first row, and $R$ be the assertion that there exists another row $r \neq 1$ with one of the largest two numbers in the first column.

#### Case 1: Both $C$ and $R$ hold
The sum of the two largest numbers in row 1 is at least $X + t$ where $t$ is in row 1 and column $c$. Thus, $a \geq X + t$. Since $t$ is one of the largest two in column $c$ and the other number is at most $X$, we get $b \leq X + t \leq a$. Similarly, $b \geq a$, so $b = a$.

#### Case 2: $R$ does not hold
If $C$ does not hold, we are done by a symmetrical argument.

Note $a_{1,t}$ is at most the second largest number of row $j$ for any $2 \leq t, j \leq n$ because $a = a_{1,1} + a_{1,t} = X + a_{1,t} = a_{j,c_j} + a_{j,d_j}$ for some $c_j, d_j$. Since $X \geq \max\{a_{j,c_j}, a_{j,d_j}\}$, $a_{1,t} \leq \min\{a_{j,c_j}, a_{j,d_j}\}$.

Consider an alternate matrix $(b_{i,j})_{2 \leq i,j \leq n}$. Initially, $b_{i,j} = a_{i,j}$. For each $t$ that satisfies $a_{1,t}$ is one of the two largest numbers in column $t$, say the two largest numbers are $a_{1,t}, a_{m,t}$. Select $2 \leq j \leq n$ such that $j \neq m$, and reset $b_{j,t}$ to be $\max\{b_{j,t}, a_{1,t}\}$. In column $t$, the two largest numbers in $(b_{i,j})$ is equal to $b$. Also, this operation doesn't change the two largest numbers in each row of the matrix $(b_{i,j})$, so the sum of the two largest in each row is $a$. We apply the inductive hypothesis on $n-1$ and $b$ to get $a = b$, as needed.

$\blacksquare$