For some integers $m,n$,proving that $p\mid m$ Assume $p$ is a prime number such that $p\equiv 1 \pmod3$, and $q=\lfloor \frac{2p}{3}\rfloor$.
If:
$$\frac{1}{1\cdot2} +\frac{1}{3\cdot4} +\cdots+\frac{1}{(q-1)\cdot q} =\frac{m}{n}$$
For some integers $m,n$, what is the proof that $p\mid m$ 

Let $p=3k+1$ and $H_n$ denote the n-th harmonic number. Since $\displaystyle \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$ (1) the sum in the question is $$\displaystyle \sum_{i=1}^q \frac{ (-1)^{i+1} }{i} = H_{2k}-H_k.$$
Working in mod $p$, we add $p$ to the denominator of each term in $H_k$ : $$ H_{2k} - H_k \equiv H_{2k} + \sum_{i=1}^k \frac{1}{p-i} = H_{p-1} =\frac{1}{2}\sum_{i=1}^{p-1} \left( \frac{1}{i} + \frac{1}{p-i}\right)\equiv \frac{1}{2}\sum_{i=1}^{p-1} \left( \frac{1}{i} + \frac{1}{-i}\right)=0.
$$
proving the result.

Using (1) we see that $$\frac{1}{1\cdot2} +\frac{1}{3\cdot4} +\cdots+\frac{1}{(q-1)\cdot q} = \frac{1}{1} - \frac{1}{2} + \frac{1}{3} \cdots  + \frac{1}{q-1} - \frac{1}{q}.$$
The series is almost like a Harmonic number but it has alternating signs, so we add and subtract 2*even terms: 
$$\frac{1}{1} - \frac{1}{2} + \frac{1}{3} \cdots  + \frac{1}{q-1} - \frac{1}{q} = \left( 1+ \frac{1}{2} + \cdots + \frac{1}{q} \right) -2 \left( \frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{q} \right).$$
The factor of 2 cancels with the denominator of every term in the second bracket. Remembering that $q=2k$ gives $$ \left( 1+ \frac{1}{2} + \cdots + \frac{1}{2k} \right) - \left( 1 + \frac{1}{2} + \cdots + \frac{1}{k} \right)= H_{2k} - H_k.$$
