Let $\triangle P_1P_2P_3$ be an equilateral triangle. For each $n\ge 4$, [i]Mingmingsan[/i] can set $P_n$ as the circumcenter or orthocenter of $\triangle P_{n-3}P_{n-2}P_{n-1}$. Find all positive integer $n$ such that [i]Mingmingsan[/i] has a strategy to make $P_n$ equals to the circumcenter of $\triangle P_1P_2P_3$.  

[i]Proposed by Li4 and Untro368.[/i]
1. **Initial Setup and Definitions:**
   - Let $\triangle P_1P_2P_3$ be an equilateral triangle.
   - For each $n \ge 4$, $P_n$ can be set as the circumcenter or orthocenter of $\triangle P_{n-3}P_{n-2}P_{n-1}$.
   - We need to find all positive integers $n$ such that $P_n$ equals the circumcenter of $\triangle P_1P_2P_3$.

2. **Strategy and Induction:**
   - We claim that the solution set is $n \equiv 0 \pmod{4}$.
   - This can be achieved by always taking $P_n$ as the orthocenter of $\triangle P_{n-3}P_{n-2}P_{n-1}$.

3. **Properties of the Triangles:**
   - By induction, $\triangle P_{n-3}P_{n-2}P_{n-1}$ will always be either a $30^\circ-30^\circ-120^\circ$ triangle or an equilateral triangle for all $n$.
   - The sides of these triangles will be parallel to those of $\triangle P_1P_2P_3$.

4. **Coordinate Representation:**
   - Let $S$ be the set of points that can be represented as:
     \[
     \left(\left(a + \frac{b}{2} \right) \cdot \frac{1}{3^n}, \left( \frac{b\sqrt{3}}{2} \right) \cdot \frac{1}{3^n} \right)
     \]
     where $a, b, n$ are integers.
   - Once a point in $S$ is fixed, the choice of $n$ does not affect the parity of $a$ and $b$ in its representation.

5. **Coloring and Parity:**
   - We color a point in $S$ based on the (constant) parity of $a$ and $b$ over all its possible representations. There are 4 different colors.

6. **Fixing the Triangle on the Coordinate Plane:**
   - Fix $\triangle P_1P_2P_3$ on the coordinate plane with:
     \[
     P_1 = (0,0), \quad P_2 = (1,0), \quad P_3 = \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)
     \]
   - $P_1, P_2, P_3$, and the circumcenter of $\triangle P_1P_2P_3$ are all in $S$ and are colored differently.

7. **Circumcenter and Orthocenter Properties:**
   - Given any equilateral triangle or $30^\circ-30^\circ-120^\circ$ triangle with vertices in $S$ and all colored differently, and with sides parallel to those of $\triangle P_1P_2P_3$, its circumcenter and orthocenter are also in $S$ and are colored in the other color.

8. **Conclusion:**
   - Only $n \equiv 0 \pmod{4}$ will yield $P_n$ that are the same color as the circumcenter of $\triangle P_1P_2P_3$.

\(\blacksquare\)

The final answer is \( \boxed{ n \equiv 0 \pmod{4} } \)