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 $\frac{1}{z_1} + \frac{1}{z_2} + \cdots + \frac{1}{z_n} = 1$?

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A related question is whether 2 can be the sum of the inverses of the positive divisors of an odd integer. But this is an open problem because it is logical equivalent to the open question whether there is an odd perfect number. – Steffen Schuler Jan 13 '13 at 6:33

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.
$$1=\frac {1}{3} + \frac {1}{5} + \frac {1}{7} + \frac {1}{9} + \frac {1}{15} + \frac {1}{21} + \frac {1}{27} + \frac {1}{35} + \frac {1}{63} + \frac {1}{105} + \frac {1}{135}$$
A similar question, are there infinitely many such nontrivial representations? By what you said, we can continually decompose each of the above fractions, but we will always get that the sum of finitely many of them will be, say $\frac{1}{3}$. Is there a representation that doesn't use $1/3$ (or any particular fraction)? – Clayton Jan 13 '13 at 4:49