Find two 2 permutations of the numbers of set $\{\frac 11,\frac 12,\dots,\frac 1n\}$ with certain properties. The problem is: Let $n>1$ be a natural numbers, $(a_1,a_2,\dots,a_n)$ and $(b_1,b_2,…,b_n)$ are $2$ permutations of the numbers in set $\{\frac 11,\frac 12,\dots,\frac 1n\}$, satisfying $a_i+b_i\ge a_j+b_j$ for all $i<j$. Does there exist $2$ permutations in which $a_i$ is not equal to $b_i$ for all $i$ and $\frac{a_1+b_1}{a_n+b_n}$ is an integer? I try for small $n$ like $3$ or $4$ and it seems to be there is no such permutation to satisfy. Can you guys help me?

Consider the counterexample:
\begin{align}
(a_1,a_2,\ldots,a_{12}) &= \bigg(\frac{1}{1},\frac{1}{4},\frac{1}{2},\frac{1}{3}, \frac{1}{5}, \frac{1}{9}, \frac{1}{6}, \frac{1}{10}, \frac{1}{7}, \frac{1}{11}, \frac{1}{8}, \frac{1}{12} \bigg) \\
(b_1,b_2,\ldots,b_{12}) &= \bigg(\frac{1}{4},\frac{1}{1},\frac{1}{3},\frac{1}{2}, \frac{1}{9}, \frac{1}{5}, \frac{1}{10}, \frac{1}{6}, \frac{1}{11}, \frac{1}{7}, \frac{1}{12}, \frac{1}{8} \bigg)
\end{align}
It is easy to see that $a_i+b_i \geqslant a_j+b_j$ for all $i<j$. Moreover:
$$\frac{a_1+b_1}{a_n+b_n} = \frac{\frac{1}{1}+\frac{1}{4}}{\frac{1}{8}+\frac{1}{12}} = \frac{\frac{5}{4}}{\frac{5}{24}} = 6$$

So what's the intuition behind this construction? We want to make sure that the numerator of $a_n+b_n$ is as small as possible to maximize our chance of getting $\frac{a_1+b_1}{a_n+b_n}$ to be an integer. Assuming one of $a_n$  and $b_n$ is $\frac{1}{n}$, we could try making the other as $\frac{1}{n/2}$. However, this will interfere with our condition that $a_i+b_i \geqslant a_j+b_j$ for all $i < j$ (I'll let you figure out why!). Thus, we could instead try keeping $\frac{1}{n}$ and $\frac{1}{2n/3}$. Then:
$$a_n+b_n = \frac{1}{n} + \frac{1}{2n/3} = \frac{5}{2n}$$
Now, we need to make sure that $5$ divides the numerator of $a_1+b_1$. It is easy to see that one of $a_1$ and $b_1$ must be $1$, so we can try taking the other as $\frac{1}{4}$. Then, we have:
$$\frac{a_1+b_1}{a_n+b_n} = \frac{\frac{1}{1}+\frac{1}{4}}{\frac{1}{2n/3}+\frac{1}{n}} = \frac{\frac{5}{4}}{\frac{5}{2n}} = \frac{n}{2}$$
We need to make sure that $\frac{n}{2}$ and $\frac{2n}{3}$ are integers, so $6 \mid n$. This strategy won't work for $n=6$ because $\frac{2n}{3} = 4$. For $n=12$, we can use the pairing trick done in the counterexample provided above to achieve the required.
