For $n=2$ we can't find such numbers. Let $d=gcd(a_1, a_2)$, $a_1 = dn$ and $a_2=dm$. Then $$\frac{a_1}{a_2} + \frac{a_2}{a_1} = \frac{n}{m} + \frac{m}{n} = \frac{n^2 + m^2}{mn} = \frac{(n+m)^2}{mn} - 2$$ should be an integer. But $n$ and $m$ are coprime, so $(n+m,n) = (n+m, m) = 1$. So if $mn|(n+m)^2$ it must be $n = m = 1$. Contradiction - we wanted distinct numbers. For $n=3$, take the sequence $(1,2,4)$. For any $n>3$, take $a_1 = a_2 = ... = a_{n-2} = 1$, $a_{n-1} = 2$, $a_{n} = 4$.

djmati11 wrote: For any $n>3$, take $a_1 = a_2 = ... = a_{n-2} = 1$, $a_{n-1} = 2$, $a_{n} = 4$. But the numbers are required to be distinct...

I assumed that "distinct" means "not all equal", not as in "pairwise distinct". Still, it's easy to fix: For odd $n$, take $a_i=2^i$. For even $n$, take $a_1 = 1, a_2 = 4, a_3 = 16$ and $a_k = 2^{k+1}$ for $4 \leq k \leq n$.