min $(x+y)$

Assuming $x,y$ are positive integers, rewrite this as $\frac1{2018} < \frac{y-x}{y} < \frac1{2017}.$ Let $a = y-x$ ; then $2017a < y < 2018a.$

If $a=1$ there is no solution. If $a=2$ there is exactly one solution, $y=4035.$ This gives $x=4033$ and $x+y = 8068.$

Larger values of $a$ yield values for $x$ and $y$ that are at least $6000,$ hence $x+y$ is at least $12000.$ So the minimum possible $x+y$ is $8068,$ so the answer is $\fbox{68}.$

Assuming problem wants $x,y$ to be positive integers...(note that $x+y$ can be an integer even if $x, y$ are not. And letting $x,y < 0$ precludes the possibility of a minimum value of $x + y$ )

Rewrite the inequality as: $\frac{4032}{4034} < \frac{x}{y} < \frac{4034}{4036}$

It is clear that $x = 4033, y = 4035$ is a solution. Didn't prove that this solution minimizes the value of $x + y$ , but verified that fact with Python routine that considered all pairs of 4-digit numbers. (I suspect a direct proof isn't that difficult, but coding is even easier.)

So $m = x+y = 8068$ .