Can the sum of finitely many inverses of distinct odd integers $\geq 3$ be 1?

Is there a positive number $n$ of distinct odd integers $z_1,z_2, \ldots, z_n \geq 3$ such that $\frac{1}{z_1} + \frac{1}{z_2} + \cdots + \frac{1}{z_n} = 1$?

In 1954, it was shown by Stewart and Breusch (independently) that if $\frac {p}{q} >0$ and $q$ is odd, then it can be written as the sum of finitely many reciprocals of odd numbers.

As a specific example,

$$1=\frac {1}{3} + \frac {1}{5} + \frac {1}{7} + \frac {1}{9} + \frac {1}{15} + \frac {1}{21} + \frac {1}{27} + \frac {1}{35} + \frac {1}{63} + \frac {1}{105} + \frac {1}{135}$$

Another solution: 1=1/3+1/5+1/7+1/9+1/11+1/13+1/23+1/721+1/979007+1/661211444787+1/622321538786143185105739+1/511768271877666618502328764212401495966764795565+1/209525411280522638000804396401925664136495425904830384693383280180439963265695525939102230139815

You may have to verify it via symbolic softwares.