We compute the sum in a finite field $\displaystyle F_{p} =\mathbb{Z} /p\mathbb{Z}$ Use bijection $\displaystyle \psi :F_{p} \longrightarrow F_{p} \ ,\ \psi ( x) =2x$ Denote the sum as $\displaystyle S=\sum ^{p-1}_{k=1}\frac{1}{k}$ Then $\displaystyle S=\sum ^{p-1}_{k=1}\frac{1}{\psi (k)} =\sum ^{p-1}_{k=1}\frac{1}{2k} =\frac{S}{2}$ ( because $\displaystyle \psi $ is also permutation of $\displaystyle \left\{\overline{1} ,\overline{2} ,\dotsc ,\overline{p-1}\right\} \ $ ) It follows that $\displaystyle S=0$. This implies $\displaystyle m$ is divisible by $\displaystyle p$.

gobathegreat wrote: If $p>2$ is prime number and $m$ and $n$ are positive integers such that $$\frac{m}{n}=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{p-1}$$Prove that $p$ divides $m$ $\color{blue}\boxed{\textbf{Proof:}}$ $\color{blue}\rule{24cm}{0.3pt}$ If $p=3:$ $$\Rightarrow \frac{m}{n}=1+\frac{1}{2}=\frac{3}{2}\Rightarrow 3|m=3_\blacksquare$$If $p>3:$ By Wolstenholme's Theorem: $$\Rightarrow p^2|m_\blacksquare$$$\color{blue}\rule{24cm}{0.3pt}$