IMO Shortlist

For any set $S$ of $n=2000$ distinct real numbers, let $D_{1} \leqslant D_{2} \leqslant \cdots \leqslant D_{m}$ be the distances between them, displayed with their multiplicities. Here $m = n(n-1)/2$. By rescaling the numbers, we may assume that the smallest distance $D_{1}$ between two elements of $S$ is $D_{1} = 1$. Let $D_{1} = 1 = y - x$ for $x, y \in S$. Evidently $D_{m} = v - u$ is the difference between the largest element $v$ and the smallest element $u$ of $S$.

If $D_{i+1}/D_{i} < 1 + 10^{-5}$ for some $i = 1, 2, \ldots, m-1$ then the required inequality holds, because $0 \leqslant D_{i+1}/D_{i} - 1 < 10^{-5}$. Otherwise, the reverse inequality
$$\frac{D_{i+1}}{D_{i}} \geqslant 1 + \frac{1}{10^{5}}$$
holds for each $i = 1, 2, \ldots, m-1$, and therefore
$$v-u = D_{m} = \frac{D_{m}}{D_{1}} = \frac{D_{m}}{D_{m-1}} \cdots \frac{D_{3}}{D_{2}} \cdot \frac{D_{2}}{D_{1}} \geqslant \left(1 + \frac{1}{10^{5}}\right)^{m-1}$$
From $m-1 = n(n-1)/2 - 1 = 1000 \cdot 1999 - 1 > 19 \cdot 10^{5}$, together with the fact that for all $n \geqslant 1$, $\left(1 + \frac{1}{n}\right)^{n} \geqslant 1 + \binom{n}{1} \cdot \frac{1}{n} = 2$, we get
$$\left(1 + \frac{1}{10^{5}}\right)^{19 \cdot 10^{5}} = \left(\left(1 + \frac{1}{10^{5}}\right)^{10^{5}}\right)^{19} \geqslant 2^{19} = 2^{9} \cdot 2^{10} > 500 \cdot 1000 > 2 \cdot 10^{5}$$
and so $v-u = D_{m} > 2 \cdot 10^{5}$.

Since the distance of $x$ to at least one of the numbers $u, v$ is at least $(u-v)/2 > 10^{5}$, we have
$$|x-z| > 10^{5}$$
for some $z \in \{u, v\}$. Since $y-x = 1$, we have either $z > y > x$ (if $z = v$) or $y > x > z$ (if $z = u$). If $z > y > x$, selecting $a = z$, $b = y$, $c = z$ and $d = x$ (so that $b \neq d$), we obtain
$$\left|\frac{a-b}{c-d} - 1\right| = \left|\frac{z-y}{z-x} - 1\right| = \left|\frac{x-y}{z-x}\right| = \frac{1}{z-x} < 10^{-5}.$$
Otherwise, if $y > x > z$, we may choose $a = y$, $b = z$, $c = x$ and $d = z$ (so that $a \neq c$), and obtain
$$\left|\frac{a-b}{c-d} - 1\right| = \left|\frac{y-z}{x-z} - 1\right| = \left|\frac{y-x}{x-z}\right| = \frac{1}{x-z} < 10^{-5}.$$
The desired result follows.