Each of integers from $1$ to $20$ are placed into the dots below. Two dots are [i]adjacent[/i], if below figure contains a line segment connecting them. Prove that how the numbers are arranged, it is possible to find an adjacent pair such that the difference between the numbers written on them is greater than $3$.

[asy]

real u=0.25cm;
for(int i = 0; i < 4; ++i) { 
real v = u*(i+1);
pair P1 = dir(90+0*72)*(0,v); 
pair P2 = dir(90+1*72)*(0,v); 
pair P3 = dir(90+2*72)*(0,v); 
pair P4 = dir(90+3*72)*(0,v); 
pair P5 = dir(90+4*72)*(0,v); 
dot(P1);dot(P2); dot(P3);dot(P4);dot(P5);
path p = P1--P2--P3--P4--P5--cycle;
draw(p);
}
[/asy]
To prove that in the given graph, it is always possible to find an adjacent pair of vertices such that the difference between the numbers written on them is greater than $3$, we can use the pigeonhole principle and properties of the graph.

1. **Graph Structure and Adjacency**:
   - The graph consists of four concentric pentagons, each with 5 vertices.
   - Each vertex in a pentagon is connected to its two adjacent vertices in the same pentagon and to the corresponding vertex in the next inner or outer pentagon.
   - This means each vertex has exactly 3 neighbors.

2. **Total Vertices and Number Assignment**:
   - There are $4 \times 5 = 20$ vertices in total.
   - We need to assign the numbers $1$ to $20$ to these vertices.

3. **Pigeonhole Principle Application**:
   - Consider any vertex $v$ with number $n$ assigned to it.
   - The neighbors of $v$ can have numbers $n_1, n_2, n_3$.
   - We need to show that at least one of the differences $|n - n_i|$ for $i = 1, 2, 3$ is greater than $3$.

4. **Minimum and Maximum Values**:
   - The smallest number is $1$ and the largest number is $20$.
   - If $n = 1$, then the neighbors must be $2, 3, 4$ to avoid a difference greater than $3$. However, this configuration is not possible for all vertices due to the total number of vertices and the need to use all numbers from $1$ to $20$.
   - Similarly, if $n = 20$, the neighbors must be $17, 18, 19$ to avoid a difference greater than $3$, which is also not possible for all vertices.

5. **General Case**:
   - For any number $n$ assigned to a vertex, the neighbors must be chosen from the remaining $19$ numbers.
   - If we try to avoid a difference greater than $3$, the neighbors must be within the range $[n-3, n+3]$.
   - This range includes $7$ numbers (if all are available), but since each vertex has 3 neighbors, and there are $20$ vertices, it is impossible to maintain this range for all vertices without overlap or repetition.

6. **Conclusion**:
   - Given the constraints and the total number of vertices, it is inevitable that at least one pair of adjacent vertices will have a difference greater than $3$.
   - This conclusion follows from the pigeonhole principle and the structure of the graph, which forces some vertices to have neighbors with a difference greater than $3$.

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\]