China Mathematical Olympiad

The answer is $0 < t < \frac{1}{2}$.

Firstly, for $0 < t < \frac{1}{2}$, choose $\lambda \in \left(0, \frac{1-2t}{2(1+t)}\right)$, let $x_i = \lambda^i$, $X = \{x_1, x_2, \dots\}$. We claim that for all (not necessarily distinct) $x, y, z \in X$, all real numbers $a$ and all positive real numbers $d$, we have the following inequality:
$$\max \{|x-(a-d)|, |y-a|, |z-(a+d)|\} > td.$$
Suppose on the contrary that there exists $a \in \mathbb{R}$, $d \in \mathbb{R}^+$ and $x_i, x_j, x_k$, such that
$$\max \{|x_i-(a-d)|, |x_j-a|, |x_k-(a+d)|\} \le td.$$
Hence
$$\begin{cases} -td \le x_i - (a-d) \le td, \\ -td \le x_j - a \le td, \\ -td \le x_k - (a+d) \le td, \end{cases}$$
i.e.
$$\begin{cases} x_i + (1-t)d \le a \le x_i + (1+t)d, \\ x_j - td \le a \le x_j + td, \\ x_k - (1+t)d \le a \le x_k - (1-t)d, \end{cases} \quad (*)$$
which implies that
$$\begin{cases} x_k - (1+t)d \le a \le x_i + (1+t)d, \\ x_i + (1-t)d \le a \le x_j + td, \\ x_j - td \le a \le x_k - (1-t)d, \end{cases}$$
note that $0 < t < \frac{1}{2}$, it follows that

$$\begin{cases} d \ge \frac{x_k - x_i}{2(1+t)}, \\ d \le \frac{x_j - x_i}{1-2t}, \\ d \le \frac{x_k - x_j}{1-2t}. \end{cases}$$
By the second and third inequalities and $d > 0$, we get $x_i < x_j < x_k$, hence $i > j > k$, $\lambda^j + \lambda^{i+1} \le \lambda^{k+1} + \lambda^i$, we get

$$\frac{x_j - x_i}{x_k - x_i} = \frac{\lambda^j - \lambda^i}{\lambda^k - \lambda^i} \le \lambda.$$
By the first and second inequalities, we get $\frac{x_j - x_i}{1-2t} \ge \frac{x_k - x_i}{2(1+t)}$, hence
$$\frac{x_j - x_i}{x_k - x_i} \ge \frac{1-2t}{2(1+t)} > \lambda,$$
which contradicts the previous inequality! Thus proved our earlier claim about $X$.

Secondly, for $t \ge \frac{1}{2}$, we show that for any infinite set $X$, for any $x < y < z$ in $X$, we can choose $a \in \mathbb{R}$ and $d \in \mathbb{R}^+$ such that
$$\max \{|x-(a-d)|, |y-a|, |z-(a+d)|\} \le td.$$
In fact, let $d = \frac{z-x}{2}$, hence $x+(1-t)d = z-(1+t)d$. Let $a = \max\{x+(1-t)d, y-td\}$. Since $t \ge \frac{1}{2}$, we obtain
$$\begin{cases} y-x < 2d \le (1+2t)d, \\ x-y < 0 \le (2t-1)d, \end{cases}$$
i.e.
$$\begin{cases} y-td \le x+(1+t)d, \\ x+(1-t)d \le y+td, \end{cases}$$
hence
$$\begin{cases} x+(1-t)d \le a \le x+(1+t)d, \\ y-td \le a \le y+td, \\ z-(1+t)d \le a \le z-(1-t)d, \end{cases}$$
from which we conclude that
$$\max\{|x-(a-d)|, |y-a|, |z-(a+d)|\} \le td.$$
So every $t \ge \frac{1}{2}$ does not satisfy the requirement of the problem.

In conclusion, the set of all required $t$ is $(0, \frac{1}{2})$.