Given an integer set, if the distances between integers in the set are still in the set, what mathematical term should be used to describe that nature? Or what nature does the set have?
For example, $x,y \in \{1,2,3,4\}$, $m=|x-y|$, $m$ still $\in \{1,2,3,4\}$. Is there a discontinuous integer set that has the nature?
To simplify notation define $\delta(x,y) = |x-y|$. Here's an easier proof (using only Euclid) of the result in the answer that Seva posted: a finite set of positive integers closed under $\delta$ a homogeneous arithmetic progression $\{d, 2d, \ldots, nd\}$.
Let $d = \gcd(A)$. By Euclid's algorithm, $\gcd(x,y)$ can be obtained from $x,y$ by repeated application of $\delta$, so $A$ is closed under pairwise $\gcd$; hence by induction $d \in A$. Then $\max(A) = nd$ for some integer $n$, and $A \subseteq \{d, 2d, \ldots, nd\}$. By repeated application of $\delta(\cdot,d)$ we find that $A$ also contains $(n-1)d$, $(n-2)d$, $(n-3)d$ etc., so $A \supseteq \{d, 2d, \ldots, nd\}$. Therefore $A = \{d, 2d, \ldots, nd\}$, QED.
It soon follows that if $A$ is an infinite set of positive integers and $A$ is closed under $\delta$ then $A$ consists of all positive multiples of $\gcd(A)$.
Assuming that your sets are finite and consist of positive integers only, every such set is a homogeneous arithmetic progression $\{d,2d,\dotsc,nd\}$, where $n=|A|$.
For the proof, let $A-A$ be the set of all differences $a_1-a_2$ with $a_1,a_2\in A$, and denote by $D_+$ the set of all positive elements of $A-A$. We have $D_+\subsetneq A$ (the inclusion is strict since the largest element of $A$ is not in $D_+$) and $|D_+|=(|A-A|-1)/2$. It follows that $|A-A|=2|D_+|+1\le 2|A|-1$, which is known to be possible if $A$ is an arithmetic progression only. It is then easy to see that the progression must be of the indicated form $\{d,2d,\dotsc,nd\}$.
Addressing a question in the comments. To derive from $|A-A|=2|A|-1$ that $A$ is an arithmetic progression, one can use induction. Write $A=\{a_1,\dotsc,a_n\}$ with $a_1<\dotsb<a_n$, and let $A_0:=A\setminus\{a_n\}$. We have $(A_0-A_0)\cup\{a_n-a_1,a_1-a_n\}\subseteq A-A$, with the union in the left-hand side disjoint. Consequently, $|A_0-A_0|\le|A-A|-2\le 2|A_0|-1$, and by the induction hypothesis, $A_0$ is an arithmetic progression. Moreover, if $a_n$ were not the next term of this progression, then $(A-A)\setminus (A_0-A_0)$ would additionally contain the elements $\pm(a_n-a_2)$, leading to $|A-A|\ge|A_0-A_0|+4>2|A|-1$, a contradiction.
Here is yet another proof; as far as simplicity is concerned (see Noam's answer), this one is difficult to beat.
Suppose that $A=\{a_1,\dotsc,a_n\}$ has the property in question, where $a_1<\dotsb <a_n$. Since $a_2-a_1$ is an element of $A$ smaller than $a_2$, we must have $a_2-a_1=a_1$; that is, $a_2=2a_1$. Similarly, $a_3-a_1$ is an element of $A$ exceeding $a_2-a_1=a_1$, but smaller than $a_3$; hence, $a_3-a_1=a_2$, implying $a_3=3a_1$. Continuing this way, we get $a_k=ka_1$ for each $k=1,2,\dotsc,n$.
