Notice that if $(x,y,z)$ is a solution then any permutation of $x,y,z$ is also a solution. We may consider such solutions as a group $S$.If $x,y,z$ are not pairwise equal,than $|S| \equiv 0 \pmod 3$ and otherwise $|S|=1$.But $2018 \equiv 2 \pmod 3$,so there exist $x_1 \neq x_2$ such that $x_i^3=x_i+m,i=1,2$.Easy calculation and $m>0$ gives $m=2$ and $x_1=2,x_2=-1$.Next we consider other solutions whose number should be 2016.Note that $2016 \equiv 0 \pmod 6$,so there are even number of groups of solutions $(x,y,z)$ with two of them equal.Assume $x=y \neq z$,then $x^2z=2x+z+2$.If $x=-1$ then $z$ can be any number in range $[-n,n]$ except $-1$.So there are $2n$ choices.If $x \neq -1$,then $z = \dfrac{2}{x-1}$,and the only solution is$(0,0,-2)$.Now the number of such solution is $2n+1$,a contradiction.

@above: for the case $x=y\neq z$ would $(3,3,1)$ also work? I think the official solution says that $(m,n) = (2,168)$ works.

stroller wrote: @above: for the case $x=y\neq z$ would $(3,3,1)$ also work? I think the official solution says that $(m,n) = (2,168)$ works. You are right.

My god ,I was too careless

Since$f(m,n)=2018$,there must exist a situation that $x=y=z$. $x^3=3x+m$ Let $x=a+b$. $LHS=(a+b)^3=a^3+b^3+3ab(a+b)=RHS$ Hence $ab=1$ ,and $a+b$ is a integer. Hence $a^3+b^3=2.m=2$ $xyz=x+y+z+2(x,y,z\neq -1)\Rightarrow \frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}=1$ And the rest is easy.