Proof or disproof of my composite conjecture. Conjecture: 
If $c_n$ is the n'th composite number,$$\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{9}+\cdots+\frac{1}{c_n}\not\in \Bbb{N}$$
for any $n$ value. 
Inspiration: I was originally intrigued by the idea of how this never works for natural numbers, and so I then thought of this question, which I have been pondering ever since. I feel, as the probability of like factors is greater, that this has a greater chance of summing to a natural number. However, the more I have thought about it, the more complex it becomes.
Attempts: I have really not understood where to start mathematically, but have used very loose logical intuition to take a jab here. I managed to concur that if there are no solutions for low $n$ values, then there is a lower chance for a solution for upper $n$ values because there is a larger chance that the composite denominator have factors of extremely obscure primes (like 17 or 31, you get the idea). 
As an answer, please prove or disprove my conjecture. A proof would be...well a proof, and a disproof might be a $n$ value that adds up to a natural number (so state the $n$). Thanks in advance. I think I am really liking it here on Stack Exchange.
 A: As,  André Nicolas's comment, that will be the best way to approach.
Let $2^k$ be the largest power of $2$ less than or equal to $c_n$. Now, consider $$\color{red}{l=lcm(2,4,6,8,9,\dots, c_n)},$$  clearly, $\color{blue}{l}$ is composite.
Now, consider the sum $$\color{red}{S=\frac l2+\frac l4+\frac l6+\frac l8+\frac l9+\dots+\dots+\frac l{2^k}+\dots+\frac l{c_n}}.$$
Look that, the fraction $\color{blue}{\frac Sl}$ equals your required sum.
Now, see that $\color{blue}{S}$ is odd, since, all the even factors of $\color{blue}{l}$ i.e. $2^k$ vanishes at the one only term where in the denominator, $\color{blue}{2^k}$ comes. And all the else terms are composite, because there all the even factors of  $\color{blue}{l}$ does not vanish.
So, ultimately, $\color{blue}{S}$ is odd and $\color{blue}{l}$ is even, so the frac $\color{blue}{\frac Sl}$ never reduces to an integer.
Hence your claim.
Note: The prime counterpart of this question is answered here.
A: It would appear that there could be a nifty proof of this adapted from the very same proof that there is only one harmonic number that is also an integer. See the following link:
https://math.stackexchange.com/a/2752/286106
Basically, go for contradiction. Assume you can stop somewhere and get an integer, and call it $m$. Find the denominator with the highest power of $2$, say $2^p$, and multiply the whole equation through by $2^{p-1}$. You'll get something like this:
$$
m = \frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\cdots+\frac{1}{c_k}
$$
$$
2^{p-1}m=2^{p-3}+\frac{n_2}{d_2}+\frac{n_3}{d_3}+\cdots+\frac{1}{2}+\cdots+\frac{n_k}{d_k}
$$
Now, I've labelled the indices to make them match up with their corresponding composite number in the sum; since the first number is $\frac{1}{4}$, it's denominator will be $d_1=1$ as long as $p\geq3$, our friend $\frac{1}{2^p}$ turns into $\frac{1}{2}$, and everyone else is stuck with an odd denominator! Now rearrange to find:
$$
-\frac{1}{2} = 2^{p-3}+\frac{n_2}{d_2}+\frac{n_3}{d_3}+\cdots-2^{p-1}m+\cdots+\frac{n_k}{d_k}
$$
But lo! the sum on the right, after reducing, must have an odd denominator, while the number on the left has an even denominator! This contradiction tells us that $m$ could not have existed in the first place unless $p=2$. Now simply check the finitely remaining cases, and we are done -- or rather, assuming I haven't made any mistakes, we have proven your conjecture.
P.S. Looks like I got beaten to the punch!
