There three types $S$: a)$S = \left\{ n \right\},n \in Z$; b)$S = \left\{ {kT + a\left| {k \in Z} \right.} \right\}$ for any positive integer $T$ and integer $a$; c) $S = \left\{ {3kT + a\left| {k \in Z} \right.} \right\} \cup \left\{ {3kT + T + a\left| {k \in Z} \right.} \right\}$ for any positive integer $T$ and integer $a$.

Thank yunxiu . 2013 China Mathematical Olympaid P1、P3： 1. Two circles $K_1$ and $K_2$ of different radii intersect at two points $A$ and $B$, let $C$ and $D$ be two points on $K_1$ and $K_2$, respectively, such that $A$ is the midpoint of the segment $CD$. The extension of $DB$ meets $K_1$ at another point $E$, the extension of $CB$ meets $K_2$ at another point $F$. Let $l_1$ and $l_2$ be the perpendicular bisectors of $CD$ and $EF$, respectively. $(1)$ Show that $l_1$ and $l_2$ have a unique common point (denoted by $P$). $(2)$ Prove that the lengths of $CA$, $AP$ and $PE$ are the side lengths of a right triangle. 3. Find all positive real numbers $t$ with the following property: there exists an infinite set $X$ of real numbers such that the inequality \[ max\{\mid x-(a-d)\mid, \mid y-a\mid, \mid z-(a+d)\mid\}>td\] holds for all (not necessarily distinct) $x,y,z\in X$, all real numbers $a$ and all positive real numbers $d$. Seehere

Lemma:For any integers $a,T$, if $a,a + T \in S$,then $3kT + a,(3k + 1)T + a \in S$ for any $k \in Z$. Proof: If $3kT + a,(3k + 1)T + a \in S$, then we have $3\left( {(3k + 1)T + a} \right) - 2\left( {3kT + a} \right) = (3k + 3)T + a = 3(k + 1)T + a \in S$① $3\left( {3kT + a} \right) - 2\left( {(3k + 1)T + a} \right) = (3k - 2)T + a = 3(k - 1)T + T + a \in S$② $3\left( {3kT + a} \right) - 2\left( {(3k - 2)T + a} \right) = (3k + 4)T + a = 3(k + 1)T + T + a \in S$③ $3\left( {(3k + 1)T + a} \right) - 2\left( {(3k + 3)T + a} \right) = (3k - 3)T + a = 3(k - 1)T + a \in S$④ From ①、③ $3(k + 1)T + a,(3(k + 1) + 1)T + a \in S$; from ②、④ we have $3(k - 1)T + a,(3(k - 1) + 1)T + a \in S$. Because for $k = 0$, $a,a + T \in S$, so we done. Solution: Obviously $S = \left\{ a \right\}$ satisfies the condition. Next we suppose $S$ has at least two elements, denotes $T = \min \left\{ {\left| {y - x} \right|\left| {x,y \in S,x \ne y} \right.} \right\}$, then there exits a integer $a$ satisfies $a,a + T \in S$. By the lemma $A = \left\{ {3kT + a,(3k + 1)T + a\left| {k \in Z} \right.} \right\} \subseteq S$. It is easy to check $S = A$ satisfies the condition. If $S \ne A$, there exits a integer $b \in S\backslash A$, then there exits a integer $r$, that $3rT + a < b < 3(r + 1)T + a$, for the minimality of $T$, $b = (3r + 2)T + a$. Because $b,b + T \in S$, by the lemma $B = \left\{ {(3k + 2)T + a\left| {k \in Z} \right.} \right\} = \left\{ {3kT + b\left| {k \in Z} \right.} \right\} \subseteq S$, so $S = A \cup B = \left\{ {kT + a\left| {k \in Z} \right.} \right\}$. Above all, there three types of $S$: a)$S = \left\{ a \right\}, a \in Z$; b)$S = \left\{ {kT + a\left| {k \in Z} \right.} \right\}$ for any positive integer $T$ and integer $a$; c) $S = \left\{ {3kT + a\left| {k \in Z} \right.} \right\} \cup \left\{ {3kT + T + a\left| {k \in Z} \right.} \right\}$ for any positive integer $T$ and integer $a$.