Three. (50 points) For a row consisting of $n$ P's, $n$ Q's, and $n$ R's, redefine a new row below it (one letter shorter). If the two letters above it are different, write the third letter in that position; if the two letters above it are the same, write that letter in that position. Repeat the operation on the newly obtained row until it becomes a single letter. Figure 5 gives an example for $n=2$.

Find all positive integers $n$ such that for any initial arrangement, after the above operations, the three vertices of the resulting triangle are either all the same or all different.
When $n=1$, there are only 6 cases as shown in Figure 10:
PQR PRQ QPR QRP RPQ RQP
$R P \quad Q P \quad R Q \quad P Q \quad Q R \quad P R$
$\begin{array}{llllll}\mathbf{Q} & \mathbf{R} & \mathbf{P} & \mathbf{R} & \mathbf{P} & \mathbf{P}\end{array}$
Figure 10

All of them meet the requirements.
For the case of $n \geqslant 2$, handle it as follows.
When $n \geqslant 2$, use $0, 1, 2$ to represent P, Q, R respectively. $x_{i, j}$ represents
the $j$-th number in the $i$-th row, then,
$$
x_{i+1, j} \equiv 2\left(x_{i, j}+x_{i, j+1}\right)(\bmod 3) .
$$

From this, the number in the last row is
$$
\begin{aligned}
x_{3 n, 1} \equiv & 2^{3 n-1}\left(x_{1,1}+C_{3 n-1}^{1} x_{1,2}+\cdots+\right. \\
& \left.C_{3 n-1}^{3 n-2} x_{1,3 n-1}+x_{1,3 n}\right)(\bmod 3) .
\end{aligned}
$$

If $n$ is a number that satisfies the condition, then for any $\left(x_{1,1}, x_{1,2}, \cdots, x_{1,3 n}\right)$, it should have
$$
31\left(x_{3 n, 1}+x_{1,1}+x_{1,3 n}\right) \text {. }
$$

And $x_{1,1}+x_{1,2}+\cdots+x_{1,3 n}$
$$
=n \times(0+1+2)=3 n \text {. }
$$

Equation (1) is equivalent to
$$
\begin{array}{c}
31\left[\left(2^{3 n-1}+1\right) x_{1,1}+2^{3 n-1} C_{3 n-1}^{1} x_{1,2}+\cdots+\right. \\
\left.2^{3 n-1} C_{3 n-1}^{3 n-2} x_{1,3 n-1}+\left(2^{3 n-1}+1\right) x_{1,3 n}\right] .
\end{array}
$$

Thus, combining with equation (2), it should have
$$
\begin{array}{l}
31\left[\left(2^{3 n-1} C_{3 n-1}^{1}-2^{3 n-1}-1\right) x_{1.2}+\cdots+\right. \\
\left.\left(2^{3 n-1} C_{3 n-1}^{3 n-2}-2^{3 n-1}-1\right) x_{1,3 n-1}\right] .
\end{array}
$$

Take $\left(x_{1.1}, x_{1.2}, \cdots, x_{1.3 n}\right)=\left(0, x_{1.2}, 1, x_{1.4}, \cdots\right.$, $\left.x_{1.3 n}\right)$ and $\left(1, x_{1.2}, 0, x_{1.4}, \cdots, x_{1.3 n}\right)$, where $x_{1.2}, x_{1.4}, \cdots$, $x_{1,3 n}$ are arbitrary, but require that $x_{1,1}, x_{1,2}, \cdots, x_{1,3 n}$ contain exactly $n$ zeros, $n$ ones, and $n$ twos. Substituting them into equation (3) and then subtracting (subtracting the numerators) yields
$$
31\left(2^{3 n-1} C_{3 n-1}^{2}-2^{3 n-1}-1\right) \text {. }
$$

Notice that
$$
\begin{array}{l}
(3 n-1)(3 n-2) \equiv 2(\bmod 3) \\
\Rightarrow \mathrm{C}_{3 n-1}^{2}=\frac{(3 n-1)(3 n-2)}{2} \equiv 1(\bmod 3),
\end{array}
$$

Thus, equation (1) becomes $31-1$, a contradiction (note that $n \geqslant 2$ is used when substituting into equation (3), where $x_{1,3}$ appears in the numerator).
Therefore, $n=1$.
(Liu Kangning, Xi'an No.1 High School, 710054)