, 2012

Assume that the numbers in the first row are $a_1 \le a_2 \le \dots \le a_n$ in some order. The numbers in the second row are $b_j = 1 - a_j$, $1 \le j \le n$. Hence $b_1 \ge b_2 \ge \dots \ge b_n$.

If $a_1 + a_2 + \dots + a_n \le \frac{n+1}{4}$, we are done (we can erase all the numbers in the second row).

Let $k$ be the least positive integer such that $a_1 + a_2 + \dots + a_k > \frac{n+1}{4}$. Then
$$a_1 + a_2 + \dots + a_{k-1} \le \frac{n+1}{4}.$$
We show that
$$b_k + b_{k+1} + \dots + b_n \le \frac{n+1}{4}.$$
Observe
$$a_k \ge \frac{a_1 + a_2 + \dots + a_k}{k} > \frac{n+1}{4k}.$$
Hence
$$b_k + b_{k+1} + \dots + b_n \le (n+1-k)b_k = (n+1-k)(1-a_k) \\ < (n+1-k) \left(1 - \frac{n+1}{4k}\right) = \frac{5}{4}(n+1) - \left(\frac{(n+1)^2 + 4k^2}{4k}\right) \\ \le (\text{AM-GM}) \frac{5}{4}(n+1) - \frac{2(n+1)(2k)}{4k} = \frac{n+1}{4}.$$