For positive integer $s$ denote by $a_s$ the number of pairs $(x, y)$, where $x, y$ are numbers on the board and $x + y = s$. Since the sum of two numbers on the board cannot be less than $2$ or greater than $2n$, the number of solutions of equation $x + y = z + w$ is $a_2^2 + \dots + a_{2n}^2$. Applying the AM-QM inequality to $a_2, \dots, a_{2n}$ gives
$$a_2^2 + \dots + a_{2n}^2 \ge \frac{(a_2 + \dots + a_{2n})^2}{2n - 1}.$$
Since each pair $(x, y)$, where $x, y$ are numbers on the board, is counted exactly once in the sum $a_2 + \dots + a_{2n}$, the sum is equal to the number of pairs $k^2$. Hence
$$\frac{(a_2 + \dots + a_{2n})^2}{2n - 1} = \frac{k^4}{2n - 1},$$
which proves the claim.