Ordering the $\binom{N}{2}$ slopes of chords of a convex function at $N$ points Recall the following :

Let $f:\mathbb{R}\to\mathbb{R}$ be a convex function, and let $x<y<z$ be distinct real numbers, then one has
  $$
\frac{f(y)-f(x)}{y-x}\leq
\frac{f(z)-f(x)}{z-x}\leq
\frac{f(z)-f(y)}{z-y}
$$
  Furthermore, if any two of these three slopes are equal, then all three are equal.

Let $N\geq 3$ be an integer, and let us identify the two element subset $\lbrace i<j\rbrace$ of $\lbrace 1,\dots,N\rbrace$ with the symbol $(ij)$.
Let $x_1<\cdots<x_N$ be $N$ distinct real numbers. Define a preorder on the set of two element subsets of $\{1,\dots,N\}$ by defining
$$
(ij)\preceq(kl)\iff s_{ij}\leq s_{kl}
$$
where, for all $1\leq i<j\leq N$
$$
s_{ij}:=\frac{f(x_j)-f(x_i)}{x_j-x_i},
$$
and $(ij) \sim (kl)$ if both $(ij)\preceq(kl)$ and $(kl)\preceq(ij)$ hold.

The above fact may be recast as follows :
  \begin{equation}
\forall~1\leq i<j<k\leq N,
\quad
(ij)\preceq(ik)\preceq(jk)
\end{equation}
  and $(ij)\sim(jk)\iff(ij)\sim(ik)\iff(jk)\sim(ik)\iff (ij)\sim(jk)\sim(ik)$.



Question 1 : Suppose $\preceq$ is a preorder on the set of all two element subsets of $\{1,\dots,N\}$ satisfying the above condition. Does there exist a convex function $f$ and $N$ points that define the same preorder by the definition outlined above ?
Question 2 : Suppose $\preceq$ can be realized with a strictly convex function $g$, can one then realize $\preceq$ using the map $f(x)=x^2$ ?

 A: As a complement to Dap's answer, here is an answer to question 2 by a slightly different method (it does not rely on equality of some slopes).
Consider six points $M_k(x_k,y_k) (1\leq k \leq 6)$ in the plane, defined as follows :
$$
\begin{array}{|c|c|c|c|c|c|c|}
\hline
k & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
x_k & 1 & 2 & 3 & 4 & 7 & 8 \\
\hline
y_k & 0 & 0 & 1 & 6 & 30 & 42 \\
\hline
\end{array}
$$
If we denote by $\sigma_k=\frac{y_{k+1}-y_k}{x_{k+1}-x_k}$ the slope from $M_k$
to $M_{k+1}$, we see that the sequence $(\sigma_k)_{1\leq k\leq 5}$ is increasing :
$$
\begin{array}{|c|c|c|c|c|c|}
\hline
k & 1 & 2 & 3 & 4 & 5  \\
\hline
\sigma_k & 0 & 1 & 5 & 8 & 12  \\
\hline
\end{array}
$$
So there is a strictly convex map $g$ such that $g(x_k)=y_k$ for $1\leq k\leq 6$. The pre-order realized by $g$ satisfies :
$$
(14) \prec (23), (25) \prec (34), (36) \prec (45), \textrm{ but } (34) \prec (16). \tag{1}
$$
This cannot be realized by $f(x)=x^2$ because in that case 
$$
s_{34}-s_{16}=\big(s_{23}-s_{14}\big)+\big(s_{34}-s_{25}\big)+\big(s_{45}-s_{36}\big) \tag{2}
$$
A: As you mention in a comment, Pascal's theorem gives an extra requirement. In particular we have the special case where the Pascal line is the line at infinity:

Given points A, B, C, D, E, F lying on a conic, if AB is parallel to DE, and BC is parallel to EF, then CD is parallel to FA.$\tag{*}$

For question 1, the answer is no at least for $N=7.$ Let $A,C,E,X,D,F,B$ denote the points $(x_1,f(x_1)),\dots,(x_7,f(x_7))$ respectively. Take the relation describing the preorder defined by a configuration meeting the hypotheses of (*) with the degenerate conic $y=|x|$ and $X=(0,0)$ - for example $x_1,\dots,x_7=-4,-2,-1,0,1,2,4$ and $f(x)=|x|.$ Then modify it by requiring that the slope of CD is greater than the slope of FA instead of equal. To satisfy this preorder, $A,C,E,X$ are forced to lie on one line, and $X,D,F,B$ are forced to lie on a different line. This doesn't destroy either of the conditions on $\preceq.$ The union of the two lines is a degenerate conic, so (*) applies and so there is no configuration of points that defines this preorder.
For question 2, the answer also no at least for $N=6.$ Let $A,C,E,D,F,B$ denote the points $(x_1,f(x_1)),\dots,(x_6,f(x_6))$ respectively, and as before take the preorder defined by a reference configuration of points on the curve $y=x^2$ meeting the hypotheses of (*), but then move $A$ slightly to the left. I haven't worked out the actual co-ordinates, but there should be a fairly simple configuration determined by $C,E,D,F=(4,-2),(1,-1),(1,1),(4,2).$ Moving $A$  modifies the preorder by requiring that that the slope of CD is greater than the slope of FA instead of equal. The requirement that the points lie on a conic means that (*) applies and so there is no configuration of points on $y=x^2$ that defines this preorder.
