Berkeley Math Circle

Solution:
Enumerate the primes $p_{1}, q_{1}, p_{2}, q_{2}, \ldots$ and define
$$a_{n}=p_{n} q_{n} \cdot \begin{cases}\prod_{k=1}^{n-2} p_{k} & n \text{ even } \\ \prod_{k=1}^{n-2} q_{k} & n \text{ odd. }\end{cases}$$
This works by construction. The idea is that you just take every pair $i<j$ you want to not be relatively prime (meaning $|i-j| \geq 2$) and throw in a prime. You can't do this by using a different prime for every pair (since each $a_{i}$ must be finite) and you can't use the same prime for a fixed $i$, so you do the next best thing and alternate using even and odd and you're done.