Summing sets of inverses of integers

[Doom of Doom]

How can I prove this?

Suppose $$X$$ is a set of 16 distinct positive integers, $$X=\left\{{x_{1}, \cdots , x_{16}}\right\}$$.
Then, for every $$X$$, there exists some integer $$k\in\left\{{1, \cdots , 8}\right\}$$ and disjoint subsets $$A,B\subset X$$
$$A=\left\{a_{1},\cdots\ ,a_{k}\right\}$$ and $$B=\left\{b_{1},\cdots\ ,b_{k}\right\}$$

such that $$\left|\alpha - \beta\right|<.00025$$,

where $$\alpha= \frac{1}{a_{1}}+\cdots+\frac{1}{a_{k}}$$ and $$\beta= \frac{1}{b_{1}}+\cdots+\frac{1}{b_{k}}$$.


I know that .00025 is pretty close to 2^-12.

 

[dodo]

Which k, A, B, would you choose for X = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16} ? It might be me, but I don't seem to find any.

 

[Gib Z]

Well for a set of 16 ordered integers, how would you choose them to get the smallest difference? Think of variance of a set of data.