About the partition of some reciprocals

Question

Does there exists a partition of the reciprocals $\frac{1}{2}, \frac{1}{3},\cdots,\frac{1}{12}$, into two sets $A$ and $B$ such that $\displaystyle\sum_A\frac{1}{n} -\sum_B\frac{1}{n}=1$?

Answer 1

Let $k=\frac{12!}{7}=1\cdot2\cdot3\cdots6\cdot8\cdots12=68428800$. Notice that $k$ is an integer and $\gcd(k,7)=1$

We could have chosen a smaller $k$, for example $\text{lcm}(1,2,3,\dots,6,8,\dots,12)=3960$, but it really doesn't change anything in the proof and it makes it clearer hopefully what is going on.

Then $k\left(\sum\limits_{A}\frac{1}{n}-\sum\limits_{B}\frac{1}{n}\right)=k$

Moving the term $\frac{k}{7}$ to one side and all other terms to the other side, we have:

$\frac{k}{7}=\pm k\pm\frac{k}{2}\pm\frac{k}{3}\pm\dots\pm\frac{k}{6}\pm\frac{k}{8}\pm\dots\pm\frac{k}{12}$

Each of the numbers on the right an integer so it follows that their sum and difference are as well. (Why?) What about the left side? What does this imply?