This is PUMAC $2008$ Division A Number Theory problem $10$.
What is the smallest number n that so that $\exists $ distinct $a_1,\cdots, a_n \in (2\mathbb{Z}_{\ge 1} + 1), \sum_{i=1}^n \dfrac{1}{a_i}=1$?
A solution I found online essentially reads as follows. One can't choose any number $n<8$ as $\dfrac{1}{a_1} + \cdots +\dfrac{1}{a_n} \leq \dfrac{1}{3}+ \dfrac{1}5 +\cdots + \dfrac{1}{15} < 1$. One can't choose 8 as one can multiply both sides by the least common multiple of the denominators to get an even number equal to an odd number. However, one can get 9 because $\sum_{i=1}^9 \dfrac{1}{a_i} = 1$ for $(a_1,a_2,\cdots, a_9) = (3,5,7,9,11,15,35,45,231).$
I was wondering if there's a theoretic way to get this tuple (i.e. without using a computer program or brute force)? I think it has something to do with inequalities and that it's good to start with $3,5,7,\cdots, 19$ as the first possible tuple. Then one needs to gradually increase numbers somehow.