First we choose distinct positive rational numbers $r_1, \dots, r_{k+3}$ such that
$$r_i r_{i+1} r_{i+2} r_{i+3} = i \quad \text{for } 1 \le i \le k$$
Let $r_1 = x, r_2 = y, r_3 = z$ be some distinct primes greater than $k$; the remaining terms satisfy $r_4 = \frac{1}{r_1 r_2 r_3}$ and $r_{i+4} = \frac{i+1}{i} r_i$. It follows that if $r_i$ are represented as irreducible fractions, the numerators are divisible by $x$ for $i \equiv 1 \pmod 4$, by $y$ for $i \equiv 2 \pmod 4$, by $z$ for $i \equiv 3 \pmod 4$ and by none for $i \equiv 0 \pmod 4$. Notice that $r_i < r_{i+4}$; thus the sequences
$$r_1 < r_5 < r_9 < \dots, \quad r_2 < r_6 < r_{10} < \dots, \quad r_3 < r_7 < r_{11} < \dots, \quad r_4 < r_8 < r_{12} < \dots$$
are increasing and have no common terms, that is, all $r_i$ are distinct.

If each $r_i$ is represented by an irreducible fraction $\frac{u_i}{v_i}$, choose a prime $p$ which divides neither $v_i$, $1 \le i \le k+1$, nor $v_i v_j$ ($r_i - r_j$) $= v_j u_i - v_i u_j$ for $i < j$, and define $a_i$ by the congruence $a_i v_i \equiv u_i \pmod{p}$. Since $r_i r_{i+1} r_{i+2} r_{i+3} = i$, we have
$$\begin{aligned} iv_i v_{i+1} v_{i+2} v_{i+3} &= r_i v_i r_{i+1} v_{i+1} r_{i+2} v_{i+2} r_{i+3} v_{i+3} \\ &= u_i u_{i+1} u_{i+2} u_{i+3} \equiv a_i v_i a_{i+1} v_{i+1} a_{i+2} v_{i+2} a_{i+3} v_{i+3} \pmod{p} \end{aligned}$$
and therefore $a_i a_{i+1} a_{i+2} a_{i+3} \equiv i \pmod{p}$ for $1 \le i \le k$.

If $a_i \equiv a_j \pmod{p}$, then $u_i v_j \equiv a_i v_j v_j \equiv u_j v_i \pmod{p}$, a contradiction. $\square$