Is it possible to write 1 as the sum of the reciprocals of x odd integers? x is the odd number. Actually, thats it. The question is:
Is it possible to write 1 as the sum of the reciprocals of x odd integers? x is the odd number. 
For  x - even the impossibility is obvious. But what to do for x - odd?
I have seen some of the similar questions, but almost everywhere there were odd numbers in a row (2n+1,n=1,2...)
And the answer was using the Harmonic number. Somewhere it is mentioned that the difference of harmonic numbers can't be an integer, but i don't know if it is true... If it is, then it is possible to write this (finite) series as a combination of harmonic numbers.
Please, correct me if i'm wrong. 
Thank you for your attention.
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 A: Also sprach Mathematica:
$$
1-1/3 - 1/5 - 1/7 - 1/9 - 1/11 - 1/15 - 1/21 - 1/135 - 1/10395=0.
$$
A: Here's a technique to generate many solutions:
Take an odd abundant number ($945$ is the smallest) and (try to) find a subset of its proper factors that adds to the number:
$7+9 + 15+27+35+45+63+105+135+189+315=945$
Divide by the original number
$\frac{7+9 + 15+27+35+45+63+105+135+189+315}{945}=1$
Write the left hand side as a sum of simplified fractions:
$\frac{1}{135}+\frac{1}{105}+\frac{1}{63}+\frac{1}{35}+\frac{1}{27}+\frac{1}{21}+\frac{1}{15}+\frac{1}{9}+\frac{1}{7}+\frac{1}{5}+\frac{1}{3}=1$
A: It's possible. I can show to you that it's possible and a constraint for generating such a sequence.
We have $k_m =$ any integer, $O_m =$ any odd number where each $O_m$ is distinct and $E_m =$ any even number. We can express it thusly:
$$\frac{k_1}{O_1} + \frac{k_2}{O_2} + \frac{k_3}{O_3} + \ ... +\frac{k_n}{O_n} = 1$$
Note that $n$ is odd.
Now multiply everything by $O_1$: $$k_1 + O_1\frac{k_2}{O_2} + O_1\frac{k_3}{O_3} + \ ... +O_1\frac{k_n}{O_n} = O_1$$
By $O_2$:
$$O_2k_1 + O_1k_2 + O_2O_1\frac{k_3}{O_3} + \ ... + \ O_2O_1\frac{k_n}{O_n} = O_1O_2$$
Continue doing this for every $O$, this will yield the following: $$ (\overbrace{O_2O_3O_4...}^{n-1})k_1 + (\overbrace{O_1O_3O_4...}^{n-1})k_2 + \ ... + \ (\overbrace{O_2O_3O_4...}^{n-1})k_n = O_2O_3O_4...O_n$$
So what happens here is that every $k_m$ gets multiplied by every $O$ except $O_m$ (because $O_m$ is already in the denominator so they cancel each other out). This will then equal the product of every $O$ times $1$ on the right hand side.
We know that $n$ is odd so $n-1$ must be even, therefore we have an even number off odd numbers multiplied together which is an odd number. This means we can rewrite the equation thusly (odd = odd):
$$k_1O + k_2O + k_3O + \ ... \ + k_nO = O_1O_2O_3...O_n$$
The right hand side $O_1O_2O_3...O_n$ is an odd number of odd integers multiplied together, which is just an odd number. So if we even want a chance for them to be the same, the left hand side needs to be odd as well. We have an odd number $n$ of terms there, and each $k$ multiplied by an odd number $O$. So there needs to be an odd number of $k$ that is odd, and an even number of $k$ that is even. This is the only way that we can get an odd number on the left hand side.
So it's possible because we don't arrive at a contradiction by setting the sum to equal $1$. I have no idea about a specific sequence like this though, but every sequence must follow the pattern I mentioned. 
EDIT First had a mistake, corrected it.
