Suppose that a numerator is greater than a denominator. Then interchanging these two numbers makes both fractions smaller. Thus we can assume w.l.o.g. that all numerators are less than all denominators.

For every $i = 1, 2, \dots, 2n$, define $x_i = a_i - i + \frac{1}{2}$. As $a_1 < a_2 < \dots < a_{2n}$ and rounding the numbers $a_i$ produce pairwise distinct positive integers, we must have $a_i \ge i - \frac{1}{2}$ which implies $x_i \ge 0$. From $a_{i+1} - a_i \le 1$, we have $x_{i+1} - x_i = a_{i+1} - a_i - 1 \le 0$ which implies $x_1 \ge x_2 \ge \dots \ge x_{2n}$.

Hence $i < j$ always implies $\frac{a_i - x_j}{a_j - x_j} \ge \frac{a_i - x_j}{a_j - x_j} \ge \frac{a_i - x_i}{a_j - x_j} = \frac{2i-1}{2j-1}$ (the first inequality holds because of $a_i < a_j$ and $x_j \ge 0$ while the second inequality holds because of $x_i \ge x_j$). Thus it suffices to prove the desired inequality for the case where the numerators are integers $1, 3, \dots, 2n - 1$ and the denominators are integers $2n + 1, 2n + 3, \dots, 4n - 1$ in some order.

Denote the sum of all fractions by $s$. Applying AM-GM to the fractions gives $\frac{s}{n} \ge \sqrt[n]{\frac{1 \cdot 3 \cdots (2n-1)}{(2n+1)(2n+3)\cdots(4n-1)}}$. Thus it suffices to prove for every $n$ the inequality $\frac{1 \cdot 3 \cdots (2n-1)}{(2n+1)(2n+3)\cdots(4n-1)} > \frac{1}{4^n}$. We can do it by induction on $n$.

The claim holds for $n = 1$ since $\frac{1}{3} > \frac{1}{4}$. For the induction step, it suffices to show that $\frac{(2n+1)^2}{(4n+1)(4n+3)} > \frac{1}{4}$ which is equivalent to $(4n+2)^2 > (4n+1)(4n+3)$. The latter follows from AM-GM for $4n+1$ and $4n+3$.

Firstly, show as in Solution 1 that it suffices to prove the desired inequality for the case where the numerators are $1, 3, \dots, 2n-1$ and the denominators are $2n+1, 2n+3, \dots, 4n-1$ in some order. By applying the rearrangement inequality for $1 < 3 < \dots < 2n-1$ and $\frac{1}{2n+1} > \frac{1}{2n+3} > \dots > \frac{1}{4n-1}$, we conclude that the least sum of fractions is obtained if both numerators and denominators are in the increasing order. Hence it suffices to show that $\frac{1}{2n+1} + \frac{3}{2n+3} + \dots + \frac{2n-1}{4n-1} > \frac{n}{4}$.

Put the first summand together with the last one, the second one together with the second last one, etc. The fractions of every pair are of the form $\frac{n-k}{3n-k}$ and $\frac{n+k}{3n+k}$ where $-n < k < n$. The sum of these two terms is $\frac{6n^2-2k^2}{9n^2-k^2}$. As $-n < k < n$, we must have $k^2 < n^2$; we also see that both $6n^2-2k^2$ and $9n^2-k^2$ are positive. Thus the inequality $9n^2-k^2 < 12n^2-4k^2$, being equivalent to the valid inequality $k^2 < n^2$, is also equivalent to $\frac{6n^2-2k^2}{9n^2-k^2} > \frac{1}{2}$. Hence the sum of members of all $n$ pairs is greater than $\frac{n}{2}$, and as each fraction occurs twice in these pairs, the sum of all fractions is greater than $\frac{n}{4}$.

Firstly, show as in Solution 1 that it suffices to prove the desired inequality for the case where the numerators are $1, 3, \dots, 2n-1$ and the denominators are $2n+1, 2n+3, \dots, 4n-1$ in some order. Replacing all denominators with their strict upper bound $4n$, all fractions become smaller and so does their sum. The sum of numerators $1+3+\dots+(2n-1)$ equals $n^2$. Hence the sum of all fractions is greater than $\frac{n^2}{4n}$, i.e., than $\frac{n}{4}$.