We claim that all $n \neq 2$ work. It's not hard to see that $\frac{1}{a_1} + \frac{1}{a_2} = 1$ has no solution in positive integers by multiplying and factoring the resulting expression. To prove that all other $n$ work, we will construct the numbers $a_1$ to $a_n$. Construction: We say that $n$ is bad if $n$ is of the form $j(j+1)$ for some positive integer $j$ and good otherwise. If $n = 1$ we simply take $a_1 = 1$. For $n > 3$, it follows from partial fraction decomposition or some other method that $\sum_{i=1}^{n-1} \frac{1}{i(i+1)}+\frac{1}{n} = 1$. If $n$ is good, we can simply set $a_i = \frac{1}{i(i+1)}$ and $a_n = \frac{1}{n}$ to get a valid construction. If $n$ is bad, then $\frac{1}{n}$ appears twice in the above construction. To solve that problem, note that $\frac{1}{6} = \frac{1}{9}+\frac{1}{18}$ and $\frac{1}{n} + \frac{1}{(n-1)n} = \frac{1}{n-1}$. Since $n \geq 6$, we may change the above construction to $a_2 = \frac{1}{9}, a_3 = \frac{1}{18}, a_i = \frac{1}{(i-1)i}$ for $4 \leq i \leq n-1$ and $a_n = \frac{1}{n-1}$. Since $n-1$ is good, the above construction satisfies the requirement.