Let $G$ be a graph whose vertices are the integers. Assume that any two integers are connected by a finite path in $G$. For two integers $x$ and $y$, we denote by $d(x, y)$ the length of the shortest path from $x$ to $y$, where the length of a path is the number of edges in it. Assume that $d(x, y) \mid x-y$ for all integers $x, y$ and define $S(G)=\{d(x, y) | x, y \in \mathbb{Z}\}$. Find all possible sets $S(G)$.
To solve the problem, we need to determine all possible sets \( S(G) \) for a graph \( G \) whose vertices are the integers, and where \( d(x, y) \mid x-y \) for all integers \( x, y \). Here, \( d(x, y) \) denotes the length of the shortest path between \( x \) and \( y \).

1. **Understanding the Problem:**
   - The graph \( G \) is connected, meaning any two integers are connected by a finite path.
   - \( d(x, y) \) is the length of the shortest path between \( x \) and \( y \).
   - \( d(x, y) \mid x-y \) implies that the distance \( d(x, y) \) divides the difference \( x-y \).

2. **Possible Sets \( S(G) \):**
   - We need to find all possible sets \( S(G) = \{d(x, y) \mid x, y \in \mathbb{Z}\} \).

3. **Constructing Examples:**
   - **Example 1:** \( S(G) = \mathbb{N} \)
     - This can be achieved with \( E = \{(x, x+1) \mid x \in \mathbb{Z}\} \). Here, every integer is connected to its successor, and the distance between any two integers \( x \) and \( y \) is \( |x-y| \), which is a natural number.
   - **Example 2:** \( S(G) = \{0, 1\} \)
     - This can be achieved with a complete graph on \( \mathbb{Z} \). In this case, every pair of distinct vertices is connected by an edge, so the distance between any two distinct integers is 1.
   - **Example 3:** \( S(G) = \{0, 1, 2\} \)
     - This can be achieved with a complete bipartite graph on \( 2\mathbb{Z} \) and \( 2\mathbb{Z} + 1 \). Here, the distance between any two integers in the same part is 2, and the distance between any two integers in different parts is 1.
   - **Example 4:** \( S(G) = \{0, 1, 2, 3\} \)
     - This can be achieved with \( E = \{(x, y) \mid x-y \equiv \pm 1 \pmod{6}\} \). Here, the distances between vertices can be 1, 2, or 3, depending on the congruence class modulo 6.

4. **Proving These Are the Only Possibilities:**
   - If \( d \in S(G) \) and \( 0 \leq d' < d \), then \( d' \in S(G) \). This is because we can take two points with distance \( d' \) on a path of length \( d \) (which connects two points at distance \( d \)).
   - Therefore, \( S(G) \) is either \( \mathbb{N} \) (i.e., the nonnegative integers) or \( [0, m] \cap \mathbb{Z} \) for some \( m \geq 1 \).

5. **Bounding \( m \):**
   - If \( m \geq 4 \), there exist two points \( x, y \) at distance 4, implying \( 4 \mid x-y \), i.e., \( x-y = 4t \) for some nonzero integer \( t \).

6. **Lemma:**
   - For all even numbers \( d \) and for all \( M > 1 \), there exist two numbers \( a, b \) such that \( (a, M) = (b, M) = 1 \) and \( a - b = d \).
   - **Proof:**
     - It suffices to find two such numbers modulo \( M \), and in particular, modulo every prime power \( p^k \) of \( M \).
     - If \( p = 2 \), choose \( 1 \) and \( 1 + d \).
     - If \( p > 2 \), one of \( (1, d+1) \) or \( (2, d+2) \) works because \( p \nmid 1, 2 \) and it also cannot divide both \( d+1 \) and \( d+2 \).

7. **Returning to the Problem:**
   - If \( M = m! \) (which is an even number) and \( (a-b, M) = 1 \), then \( d(a, b) \mid m! \) and \( d(a, b) \mid a-b \implies d(a, b) = 1 \).
   - Choosing two numbers \( a, b \) such that \( a - b = 4t \) and \( (a, M) = (b, M) = 1 \), we can choose the point \( z = x + b = y + a \) which has difference \( b \) from \( x \) and \( a \) from \( y \).
   - Therefore, \( d(z, x) = 1 \) and \( d(y, z) = 1 \), which implies \( d(x, y) \leq d(x, z) + d(z, y) = 2 \), contradicting \( d(x, y) = 4 \).

Thus, the only possible sets \( S(G) \) are \( \mathbb{N} \), \( \{0, 1\} \), \( \{0, 1, 2\} \), and \( \{0, 1, 2, 3\} \).

The final answer is \( \boxed{ \mathbb{N}, \{0, 1\}, \{0, 1, 2\}, \{0, 1, 2, 3\} } \).