Tricky fraction

$\frac{1}{2016}>\frac{m}{n}>\frac{1}{2017}$

We have to take minimum values.So multiplying $2$ in neumerator and denominator :

$\frac{2}{4032}>\frac{m}{n}>\frac{2}{4034}$

The smallest $\frac{m}{n}$ is $\dfrac{2}{4033}$

The simplest way is bring n to numerator, the first term should be greater than 2, middle 2 and first less than 2. So n=4033 and m=2 satisfy minimum...

In general, we can use the Farey Sequence to solve this question. This question is equivalent to finding the smallest denominator between the two fractions (since the numerator would be minimised as well, so their sum is minimised). Note that if $|ad-bc|=1$ , then, these two terms are adjacent terms in the Farey sequence. Thus, the fraction with the smallest denominator between them is

$\dfrac {a}{b} < \dfrac {a+c}{b+d} < \dfrac {c}{d}$

Therefore, if $|ad-bc=1|$ , then the fraction with the smallest denominator between them is $\dfrac {a+c}{b+d}$ , so the general sum is $a+b+c+d$ . Substituting the values is the question, we find that the if condition is satisfied, so the answer is 4035.