Assume $n - m \geq 2$. Then $f(n) = f(f(m)) = 2f(m) - m - 2 = 2n-m-2 \geq n$, so by taking $m'=n$, $n'= 2n-m-2$, we have $n' - m' = (n-m) - 2 \geq 0$ and $f(m') = n'$, and the conditions are repeated. Therefore we may reduce to the case when $m\leq n\leq m+1$. Now, if $n=m$, then $m = f(n) = f(f(m)) = 2f(m) - m - 2 = m-2$, absurd. If $n=m+1$, then $f(m+1) = f(f(m)) = 2f(m) - m - 2 = m$, and then $m+1 = f(m) = f(f(m+1)) = 2f(m+1) - (m + 1) - 2 = m-3$, absurd. So no such function $f$ and pair $m,n$ will exist. In principle, one should exhibit a function satisfying i), without the ii) condition, so as to show that condition ii) is instrumental. My post here http://www.artofproblemsolving.com/Forum/viewtopic.php?f=36&t=460567 asked precisely for this; I am told that, on another math forum, Iurie Boreico indeed showed a way to build such a function.

let $n-m=k-1$ $k$ is a positive integer $f(m)=n$,$f(n)=2n-m-2$,$f(2n-m-2)=3n-2m-6$ and by induction, $f(kn-(k-1)m-k(k-1))=(k+1)n-km-(k+1)k$,here $kn-(k-1)m-k(k-1)=m$,but: $(k+1)n-km-(k+1)k-n$ is not equal to $n$ . a contradiction.

We claim that there do not exist such an $m,n$ and $f$. from $ii)$ condition we know that for integer $m \leqslant n$ , $f(m)=n$. so we claim that there do not exist such integers $m$ and $r$ s.t. $f(m)=m+r$ for $r \geqslant 0$ we observe that for $r=0$ claim is true as $f(f(m))=2f(m)-m-2=2m-m-2=m-2=m$ which is not true for any integer $m$. for $r=1$ also claim is true as $f(f(m))=2f(m)-m-2=m$ so $f(m+1)=m \implies f(f(m+1))=f(m)=m+1=2f(m+1)-m-3 \implies m+1=m-3$ again contradicition for any integer $m$. Now , suppose $f(m)=m+r$ for some $r \geqslant 2$ for such a minimal $r$ , we have $f(m+r)=f(f(m))=2f(m)-m-2 \implies f(m+r)=(m+r)+(r-2)$,take $m+r=m'$ and $r-2=r'$ which gives $f(m')=m'+r'$ which contradicts the minimality of $r$ for $r \geqslant 2$ , which forces that there are no such integers $m$ and $r$ with such a condition , and hence the claim follows. $\square$ so there do not exists such $f,m,n$. $\blacksquare$