, 2022

The answer is all multiples of 4.

Let $O$ be the circumcenter of $\Delta P_1 P_2 P_3$. Then as long as Xiao Ming always chooses the orthocenter, we have $P_{4k} = O, P_{4k+1} = P_1, P_{4k+2} = P_2$ and $P_{4k+3} = P_3$, for all $k \in \mathbb{N}$. Therefore all $n$ divisible by 4 satisfy the requirement.

Next we prove that $n$ must be divisible by 4. This requires the following two lemmas.

Lemma 1. For all $n \ge 3$, $\Delta P_{n-2} P_{n-1} P_n$ is either an isosceles triangle with a vertex angle of 120° (called type A) or an equilateral triangle (called type B).

Proof: Let us proceed by mathematical induction on $n \ge 3$. When $n=3$, it is clearly type B. When $n>3$:
- When $\Delta P_{n-3} P_{n-2} P_{n-1}$ is type A: if Xiao Ming chooses the circumcenter and $P_{n-3}$ is the vertex angle, then $\Delta P_{n-2} P_{n-1} P_n$ is type A; if Xiao Ming chooses the circumcenter and $P_{n-2}$ or $P_{n-1}$ is the vertex angle, then $\Delta P_{n-2} P_{n-1} P_n$ is type B.
If Xiao Ming chooses the orthocenter and $P_{n-2}$ or $P_{n-1}$ is the vertex angle, then $\Delta P_{n-2} P_{n-1} P_n$ is type A; if Xiao Ming chooses the orthocenter and $P_{n-3}$ is the vertex angle, then $\Delta P_{n-2} P_{n-1} P_n$ is type B.
- When $\Delta P_{n-3} P_{n-2} P_{n-1}$ is type B, regardless of whether Xiao Ming chooses the circumcenter or the orthocenter, $\Delta P_{n-2} P_{n-1} P_n$ is always type A.
By mathematical induction, the lemma is proved. □

In what follows, without loss of generality, assume $O$ is the origin. We consider oblique coordinates $(x, y)$, representing the point $x\vec{P}_1 + y\vec{P}_2$. In this case, $P_3 = (-1, -1)$.

Lemma 2. Every point that Xiao Ming may choose must have the form $(\frac{a}{3^r}, \frac{b}{3^s})$, where $a, b$ are integers coprime to 3, and $r, s$ are non-negative integers. Furthermore, for all $n \ge 4$, the coordinates of $P_{n-3}, P_{n-2}, P_{n-1}, P_n$ are all distinct modulo 2 (here we take $3^{-1} \equiv 1$), that is, $(0,0), (0,1), (1,0), (1,1)$ each appear exactly once.

Proof: We use mathematical induction on $n$. Since $P_4 = O$, the case $n=4$ holds. When $n>4$, rename $P, Q, R$ as the vertices of $\Delta P_{n-3} P_{n-2} P_{n-1}$, such that $P$ is the one with the largest angle. By Lemma 1 and its proof, we know that $P_n$ must be one of $\vec{Q} + \vec{R} - \vec{P}, 3\vec{P} - \vec{Q} - \vec{R}, \frac{1}{3}(\vec{P} + \vec{Q} + \vec{R})$. Since the coefficients involve division by at most 3, we can easily see that the denominator of each coordinate component must be a power of 3.

$$P_n + P + Q + R \equiv (0,0) \pmod{2}.$$
By the induction hypothesis, $P_{n-3}$, $P_{n-2}$, $P_{n-1}$ are all distinct, so from
$$(0,0) + (0,1) + (1,0) + (1,1) \equiv (0,0) \pmod{2}$$
we know that the coordinates of $P_{n-3}$, $P_{n-2}$, $P_{n-1}$, $P_n$ are all distinct modulo 2. □

Now, returning to the original problem. By Lemma 2, we have
$$P_{4k+1} \equiv P_1 \equiv (1,0), \quad P_{4k+2} \equiv P_2 \equiv (0,1),$$
$$P_{4k+3} \equiv P_3 \equiv (1,1), \quad P_{4k} \equiv P_4 = O \equiv (0,0),$$
therefore only multiples of 4 can achieve the requirement of the problem.