Let $n$ be a positive integer and $P_1, P_2, \ldots, P_n$ be different points on the plane such that distances between them are all integers. Furthermore, we know that the distances $P_iP_1, P_iP_2, \ldots, P_iP_n$ forms the same sequence  for all $i=1,2, \ldots, n$ when these numbers are arranged in a non-decreasing order. Find all possible values of $n$.
1. **Claim**: The only possible values of \( n \) are \( 1, 2, 3, 4, 6 \).

2. **Verification for small \( n \)**:
   - For \( n = 1 \), there is only one point, so the condition is trivially satisfied.
   - For \( n = 2 \), two points can be placed at any distance apart, satisfying the condition.
   - For \( n = 3 \), consider the vertices of an equilateral triangle with side length 1. The distances between any two points are the same, satisfying the condition.
   - For \( n = 4 \), consider the vertices of a \( 3 \times 4 \) rectangle. The distances between the points are integers and form the same sequence for each point.
   - For \( n = 6 \), construct an equilateral triangle \( V_1V_3V_5 \) with side length 7. Rotate it about its center to form another triangle \( V_1'V_3'V_5' \) such that \( V_1V_1' = V_3V_3' = V_5V_5' = 3 \). The points \( V_1, V_3, V_5, V_1', V_3', V_5' \) form a suitable configuration.

3. **Lemma 1**: Any set of points satisfying the property in the question must be concyclic.
   - **Proof**: This is a known result in geometry. If the distances between points are the same when arranged in non-decreasing order, the points must lie on a common circle. This is because the distances from any point to all other points must be the same, implying a circular arrangement.

4. **Case Analysis**:
   - **Odd \( n \)**: If \( n \) is odd, each distance must appear twice per vertex. This implies that the polygon formed by the vertices is regular. Therefore, only \( n = 1, 3 \) work for odd \( n \).
   - **Even \( n \)**: If \( n \) is even, the points must form a union of two concyclic \( \frac{n}{2} \)-gons. For \( n = 2, 4 \), this is straightforward. For \( n > 4 \), consider the smallest distance. If it appears twice per vertex, the polygon is an \( n \)-gon, which is a union of two concyclic \( \frac{n}{2} \)-gons. If it appears only once, label the vertices \( v_1, v_2, \ldots, v_n \) in clockwise order such that \( v_{2i-1}v_{2i} \) is the minimal distance. The distances from \( v_1 \) to \( v_1, v_3, \ldots, v_{n-1} \) are the same as from \( v_2 \) to \( v_2, v_4, \ldots, v_n \). This implies that \( v_1, v_n \) are the same distance apart as \( v_2, v_3 \). Using similar arguments, the result is proven.

5. **Irrationality Argument**:
   - For \( n > 3 \), \( n \)-gons cannot have all sides and diagonals of integral length because \( \cos \frac{2 \pi}{n} \) is irrational. This shows that even \( n \ge 8 \) do not work.

6. **Conclusion**: The only possible values of \( n \) are \( 1, 2, 3, 4, 6 \).

The final answer is \( \boxed{1, 2, 3, 4, 6} \).