A fun problem. The answer is $3/4$.

**Lower bound**: We first make some relaxations to the problem.

Firstly we observe that we can relax the requirement that $f$ be bijective to $f$ being injective, because any injective map can converted to be a bijection by modifying it on an infinite set $A$ of density zero (replacing $f$ on $A$ with some bijection between the equinumerable sets $A$ and $f(A) \cup f({\bf N})^c$), which does not affect $\mathrm{rc}(f)$.

Secondly we can relax the requirement that $f$ be injective further, to the requirement that $f$ is proper (all preimages of $f$ are finite). Indeed, if we have a proper map $f$, enumerate each preimage $f^{-1}(k)$ in increasing order as $N_{k,1}<\dots< N_{k,a_k}$ for some $a_k \geq 0$, and then one can check that the modified map $\tilde f: {\bf N} \to {\bf N}$ defined by
$$ \tilde f( N_{k,i} ) := (\sum_{j=1}^k a_j) - i + 1$$
is injective with $\mathrm{rc}(f) = \mathrm{rc}(\tilde f)$ (where in order to extend $\mathrm{rc}$ to non-injective $f$ one should use the closed condition $(m-n)(f(m)-f(n)) \leq 0$ rather than the open condition $(m-n)(f(m)-f(n)) < 0$). (Actually this second reduction subsumes the first, since $\tilde f$ is in fact bijective, but I retain the first reduction as I think it is a little more intuitive, and was the one I discovered first.)

Heuristically, the problem becomes easier the more slowly growing we allow $f$ to be, as this creates lots of collisions $f(n)=f(m)$ that automatically contribute to $\mathrm{rc}(f)$. A monotonic slowly growing function, such as $f(n) = \lfloor \log \log (n+100) \rfloor$, is proper, but only gives the lower bound of $1/2$ mentioned previously, with the main contribution coming from scales where $f$ transitions abruptly from one value $i$ to the next. To reduce the effect of this transition we perform a random construction to "smooth things out". Let $g: {\mathbf R}^+ \to {\mathbf R}^+$ be a sufficiently slowly growing function, e.g., $g(x) = \log\log(x+100)$ will suffice, and define a random function $f : {\bf N} \to {\bf N}$ by declaring $f(n)$ to equal $\lfloor g(n) \rfloor$ with probability $1 - \{g(n)\}$ and $\lfloor g(n) \rfloor+1$ with probability $\{g(n)\}$, independently in $n$. [One can think of $f$ as a "pointillist" discretization of $g$.] Clearly $f$ is proper, and it is a routine matter (using the Chernoff inequality) to verify that $\mathrm{rc}(f)=3/4$ almost surely if $g$ is sufficiently slowly growing (the key scales $N$ are those where $g(N)$ is close to a half-integer). [Compare with the indicator function $1_A$ of a random set of density $p$, where one can compute $\mathrm{rc}(1_A) = 1 - p(1-p) \geq 3/4$ almost surely, with equality at $p=1/2$, although this function is not directly eligible for consideration as it will not be proper.]

**Upper bound**: For each $N$, let $L_N$ be the median value of $\{f(1),\dots,f(N)\}$. As observed previously, these medians go to infinity as $N \to \infty$. Thus we can find arbitrarily large $N$ which have "world record medians" in the sense that $L_n < L_N$ for all $1 \leq n \leq N$. (In the lower bound example, these correspond to scales where $g(N)$ is close to a half-integer, which we previously identified to be critical.) We remark that this observation only requires $f$ to be proper, rather than bijective.

Let $N$ be as above, and let $E_N$ be the set of all $1 \leq n \leq N$ such that $f(n) \leq L_N$, then $E_N$ has cardinality $N/2+O(1)$; furthermore, for any $n \leq N$, $E_N \cap \{1,\dots,n\}$ has cardinality at least $n/2+O(1)$. The quantity
$$ |\{(m,n) \in \{1,\dots,N\}^2: (m-n) (f(m)-f(n)) > 0 \}|$$
is equal by symmetry to
$$ 2 |\{ (m,n): 1 \leq m < n \leq N; f(n) > f(m) \}|$$
which is at least
$$ 2 |\{ (m,n): 1 \leq m < n \leq N; m \in E_N, n \not \in E_N \}|$$
which we can write as the difference of
$$ 2 |\{ (m,n): 1 \leq m < n \leq N; m \in E_N \}| = 2 \sum_{n=1}^N |E_N \cap \{1,\dots,n-1\}|$$
and
$$ 2 |\{ (m,n): 1 \leq m < n \leq N; n, m \in E_N \}| = 2 \binom{|E_N|}{2}.$$
By hypothesis, the first quantity is at least
$$ 2 \sum_{n=1}^N (n/2+O(1)) = N^2/2 + O(N)$$
and the second quantity is $N^2/4 + O(N)$. Thus
$$ |\{(m,n) \in \{1,\dots,N\}^2: (m-n) (f(m)-f(n)) > 0 \}| \geq N^2/4 + O(N)$$
or equivalently
$$ |\{(m,n) \in \{1,\dots,N\}^2: (m-n) (f(m)-f(n)) < 0 \}| \leq 3N^2/4 + O(N)$$
giving the upper bound.