Maybe the easiest of all time CMO 3rd questions. Firstly, we show that for any infinite set X, there exists $x,y,z\in X$, $d>0, a$ s.t. \[ max\{\mid x-(a-d)\mid,\mid y-a\mid,\mid z-(a+d)\mid\}=\frac{1}{2}d\] We can choose $z=y,x<y, d=y-x, a=\frac{x+y}{2}$ and see that $\mid x-(a-d)\mid=\mid y-a\mid=\mid z-(a+d)\mid=\frac{1}{2}d$. Secondly, we show that for any $0<t<\frac{1}{2}$, if we choose $s=max(\lceil{log_{(1-2t)}\frac{1}{2}}\rceil,0)+2,X=\{-(1-2t)^{sk}\}$(k is positive integers) , then for any $ x,y,z\in X $, $d>0,a$ \[ max\{\mid x-(a-d)\mid,\mid y-a\mid,\mid z-(a+d)\mid\}> td\]. 1)if $x=y$, then $\mid x-(a-d)\mid+\mid y-a\mid\ge\mid x-a+d+a-x\mid=d$, so $max\{\mid x-(a-d)\mid,\mid y-a\mid,\mid z-(a+d)\mid\}\ge \frac{1}{2}d$ 2)if $x=z$, then $\mid x-(a-d)\mid+\mid z-(a+d)\mid\ge x-a+d+a+d-x=2d$, so $max\{\mid x-(a-d)\mid,\mid y-a\mid,\mid z-(a+d)\mid\}\ge d$ 3)if $y=z$, then $\mid y-a\mid+\mid z-(a+d)\mid\ge y-a+a+d-y=d$, so $max\{\mid x-(a-d)\mid,\mid y-a\mid,\mid z-(a+d)\mid\}\ge \frac{1}{2}d$ 4)$x,y,z$ are distinct. Suppose the contrary, then $\mid x-(a-d)\mid \le td$,$\mid y-a\mid \le td$,$\mid z-(a+d)\mid \le td$, we can conclude that $a+(-1-t)d\le x \le a+(t-1)d,a-td \le y \le a+td, a+(1-t)d \le z \le a+(1+t)d$. This is impossible, since $z-y \ge(1-2t)d,y-x \le (2t+1)d$,$\frac{y-x}{z-y} \le \frac{2t+1}{1-2t}$, on the other hand, $\frac{y-x}{z-y}>\frac{1-(1-2t)^s}{(1-2t)^s}>\frac{2t+1}{1-2t}$, a contradiction.

Ok,I posted my solution(Or only ideas) because some friends asked for mine. Good students in Cmo thought this problem is not a very hard problem,and someone said this is a mathematical analysis problem.But I think it is a linear programming problem(线性规划问题). First for $t=\frac{1}{2}$,the example is not hard,because for any $x_1<x_2,x_1,x_2\in X$,you can choose $a\in \mathbb{R},d>0$,such that:$x_2=a-\frac{d}{2},x_2=a+d$,then set $x=x_1,y=x_1,z=x_2$ doesn't satisfy the condition. For $0<t<\frac{1}{2}$,I will construct an example of inifinte set $X$ by induction. Note that if $x,y,z$ doesn't satisfy the condition,we have: \[(-1-t)d+a\le x \le (-1+t)d+a,-td+a\le y \le td+a,(1-t)d+a\le z \le (1+t)d+a.\] Then we have $x< y < z$.Now if you find have found a finite set $X_n=\{x_1,x_2,\cdots,x_n\},x_1<x_2<\cdots<\cdots<x_n$,you have to find a real number $x_{n+1}>x_n$ to such that $X_{n+1}=\{x_1,x_2,\cdots,x_n,x_{n+1}\}$ still satisfies the condition.Now draw an axis $dOa$(the $d$ is $X$-axis and the $a$ is the $Y$-axis.) Now you just choose the $x_{n+1}$ satisfies the area $\Omega$ defined by \[d>0,(1-t)d+a\le x_{n+1} \le (1+t)d+a\] has no intersection with the area $\Omega_{i,j}$ defined by \[d>0,(-1-t)d+a\le x_i \le (-1+t)d+a,-td+a\le x_j \le td+a\] But with the picture you draw you will find there is a number $a$,such that if $x_{n+1}>a$,then $\Omega\cap \Omega_{1,n}$.And Indeed,such $x_{n+1}$ satisfies the condition. I am sorry,I haven't wrote an formal solution,I just wrote an approach.

The ansewr is all $t<\frac{1}{2}$. Firstly, we show that for all $x<y<z\in \mathbb R$ there exists $a$ real and $d$ positive such that $$\max\{|x-(a-d)|,|y-a|,|z-a-d|\}\leq \frac{1}{2}d$$Indeed, WLOG assume $x-y<y-z$, let $c=\frac{1}{4}(x+y-2z)$, then there exists $a,d$ such that \begin{align*} z-c&=a+d\\ y+c&=a\\ x-c&=a-d \end{align*}Let $$k=\frac{y-x}{x+y-2z}$$Then $$\frac{c}{d}=\frac{1}{2+4k}<\frac{1}{2}$$ Conversely, we may choose $X=\{1,m,m^2,...,m^n,..\}$, where $m$ is very large. Suppose $m^{n_1},m^{n_2},m^{n_3}$ are elements of $X$. Suppose $a_1,a_2,a_3$ is an arithmetic sequence, define $c_i=m^{n_i}-a_i$, then $$m^{n_1}+m^{n_3}-2m^{n_2}=c_1+c_3-2c_2\leq |c_1|+|c_3|+2|c_2|$$$$d=a_2-a_1=m^{n_2}-c_2-m^{n_1}+c_1\leq m^{n_2}-m^{n_1}+|c_2|+|c_1|$$Hence $$\frac{|c_1|+|c_3|}{d}\geq \frac{|c_1|+|c_3|}{m^{n_2}-m^{n_1}+|c_2|+c_1|}\geq \frac{|c_1|+|c_2|}{(2k+1)|c_2|+(k+1)|c_1|+k|c_3|}\geq \frac{1}{2k+1}$$By pigeonhole principle the maximum of $$\{\frac{|c_1|}{d},\frac{|c_3|}{d}\}$$can be made arbitrarily colosed to $\frac{1}{2}$.