Let $\frac{a}{b} = 1-\frac{1}{2}+\frac{1}{3}-\dots+\frac{1}{67},$ such that $\gcd(a,b)=1$. Show that $101\mid a.$ Let $\frac{a}{b} = 1-\frac{1}{2}+\frac{1}{3}-\dots+\frac{1}{67},$ such that $\gcd(a,b)=1$. Show that $101\mid a.$ 
i got this problem in olympiad mathematics .   
$\frac{a}{b}=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{67}-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{66}\right)$
$\frac{a}{b}=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{67}-\left(\frac{1}{1}+\frac{1}{2}+...+\frac{1}{33}\right)$
$\frac{a}{b}=\frac{1}{34}+\frac{1}{35}+...+\frac{1}{67}$
Then what will I do?

Hint:
You're on the right track. In the new sum you have, you can pair up the terms $1/n$ and $1/(101-n)$:
$$\sum_{n=34}^{50} \frac{1}{n}+\frac{1}{101-n}.$$
Can you take it from here?
