Minimize $m+n$ given $\dfrac{2016}{2017}<\dfrac mn<\dfrac{2017}{2018}$

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

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

$$2016<\frac{n}{m-n}<2017$$

So $n\ge 2\cdot2016+1=4033$ and $m-n\ge 2$. Hence $m\ge 4035$

The smallest value is $8068$.

I believe that you will have to use the mediant operator, which does the following to two fractions in simplest form: $$\frac{a}{b}\oplus \frac{c}{d}=\frac{a+c}{b+d}$$ if you apply this to your fractions, you will get $$\frac{2016}{2017}\oplus \frac{2017}{2018}=\frac{4033}{4035}$$ And the answer should be $4033+4035=8068$. The mediant gives the fraction that is strictly between two other fractions, and I believe that it minimizes the numerator and denominator while doing so. To read more about the mediant, see this page.

Hint: $$\frac{a}{b} < \frac{a+c}{b+d} < \frac{c}{d}$$