Problem 2096

For $k=1$, there is nothing to prove. Henceforth assume $k \geq 2$.
Let $p_{1}, p_{2}, \ldots, p_{k}$ be $k$ distinct primes such that
$$k < p_{k} < \cdots < p_{2} < p_{1}$$
and let $N = p_{1} p_{2} \cdots p_{k}$. By Chinese Remainder Theorem, there exists a positive integer $x$ satisfying
$$x \equiv -i \quad (\bmod\ p_{i})$$
for all $i=1,2, \ldots, k$ and $x > N^{2}$. Consider the following sequence:
$$\frac{x+1}{N}, \quad \frac{x+2}{N}, \quad \ldots, \quad \frac{x+k}{N}.$$
This sequence is obviously an arithmetic sequence of positive rational numbers of length $k$. For each $i=1,2, \ldots, k$, the numerator $x+i$ is divisible by $p_{i}$ but not by $p_{j}$ for $j \neq i$, for otherwise $p_{j}$ divides $|i-j|$, which is not possible because $p_{j} > k > |i-j|$. Let
$$a_{i} := \frac{x+i}{p_{i}}, \quad b_{i} := \frac{N}{p_{i}} \quad \text{ for all } i=1,2, \ldots, k$$
Then
$$\frac{x+i}{N} = \frac{a_{i}}{b_{i}}, \quad \operatorname{gcd}(a_{i}, b_{i}) = 1 \quad \text{ for all } i=1,2, \ldots, k$$
and all $b_{i}$'s are distinct from each other. Moreover, $x > N^{2}$ implies
$$a_{i} = \frac{x+i}{p_{i}} > \frac{N^{2}}{p_{i}} > N > \frac{N}{p_{j}} = b_{j} \quad \text{ for all } i, j = 1,2, \ldots, k$$
and hence all $a_{i}$'s are distinct from $b_{i}$'s. It only remains to show that all $a_{i}$'s are distinct from each other. This follows from
$$a_{j} = \frac{x+j}{p_{j}} > \frac{x+i}{p_{j}} > \frac{x+i}{p_{i}} = a_{i} \quad \text{ for all } i < j$$
by our choice of $p_{1}, p_{2}, \ldots, p_{k}$. Thus, the arithmetic sequence
$$\frac{a_{1}}{b_{1}}, \quad \frac{a_{2}}{b_{2}}, \quad \ldots, \quad \frac{a_{k}}{b_{k}}$$
of positive rational numbers satisfies the conditions of the problem.

For any positive integer $k \geq 2$, consider the sequence
$$\frac{(k!)^{2}+1}{k!}, \frac{(k!)^{2}+2}{k!}, \ldots, \frac{(k!)^{2}+k}{k!}$$
Note that $\operatorname{gcd}(k!, (k!)^{2}+i) = i$ for all $i=1,2, \ldots, k$. So, taking
$$a_{i} := \frac{(k!)^{2}+i}{i}, \quad b_{i} := \frac{k!}{i} \quad \text{ for all } i=1,2, \ldots, k$$
we have $\operatorname{gcd}(a_{i}, b_{i}) = 1$ and
$$a_{i} = \frac{(k!)^{2}+i}{i} > a_{j} = \frac{(k!)^{2}+j}{j} > b_{i} = \frac{k!}{i} > b_{j} = \frac{k!}{j}$$
for any $1 \leq i < j \leq k$. Therefore this sequence satisfies every condition given in the problem.