$$
\begin{array}{l}
\text { 2 For a row consisting of } n \text { A's, } n \text { B's, and } n \text { C's, define a new row below it (one letter shorter) such that if the two letters above it are different, write the third letter, and if they are the same, write that letter. Repeat this operation on the newly obtained row until only one letter remains. The table below gives an example for } n=2. \\
\text { B A A A C } \\
\text { C A A B } \\
\text { B A C } \\
\text { C B } \\
\text { A } \\
\end{array}
$$

Find all positive integers \( n \) such that for any initial arrangement, after the above operation, the three vertices of the resulting large triangle are either all the same or all different.
Solution: There is only one solution: $n=1$.

Method 1: When $n=1$, there are only the following 6 cases:
\begin{tabular}{cccccc}
$\mathrm{ABC}$ & $\mathrm{ACB}$ & $\mathrm{BAC}$ & $\mathrm{BCA}$ & $\mathrm{CAB}$ & $\mathrm{CBA}$ \\
$\mathrm{CA}$ & $\mathrm{BA}$ & $\mathrm{CB}$ & $\mathrm{AB}$ & $\mathrm{BC}$ & $\mathrm{AC}$ \\
$\mathrm{B}$ & $\mathrm{C}$ & $\mathrm{A}$ & $\mathrm{C}$ & $\mathrm{A}$ & $\mathrm{B}$
\end{tabular}

All of these meet the requirements.
When $n \geqslant 2$, consider the following permutation:
$$
\text { A C B A B C } \cdots \text { A B C, }
$$

where $\mathrm{ACB}, \mathrm{ABC}$ repeat.
Consider the operation of the subarrays formed by any 4 consecutive elements in the above permutation, which can be as follows:
\begin{tabular}{cccccc}
$\mathrm{ABCA}$ & $\mathrm{BCAC}$ & $\mathrm{CACB}$ & $\mathrm{ACBA}$ & $\mathrm{CBAB}$ & $\mathrm{BABC}$ \\
$\mathrm{CAB}$ & $\mathrm{ABB}$ & $\mathrm{BBA}$ & $\mathrm{BAC}$ & $\mathrm{ACC}$ & $\mathrm{CCA}$ \\
$\mathrm{BC}$ & $\mathrm{CB}$ & $\mathrm{BC}$ & $\mathrm{CB}$ & $\mathrm{BC}$ & $\mathrm{CB}$ \\
$\mathrm{A}$ & $\mathrm{A}$ & $\mathrm{A}$ & $\mathrm{A}$ & $\mathrm{A}$ & $\mathrm{A}$
\end{tabular}

This indicates: In the listed table, all the letters in the 4th row are $\mathrm{A}$. Therefore, the number in the last row is $\mathrm{A}$, which does not meet the requirement.
Thus, the only $n$ that satisfies the condition is $n=1$.

Method 2: The case for $n=1$ is handled as above. For the case $n \geqslant 2$, handle it as follows.

When $n \geqslant 2$, use $0, 1, 2$ to represent $\mathrm{A}, \mathrm{B}, \mathrm{C}$ respectively; $x_{i, j}$ represents the number in the $j$-th column of the $i$-th row, then $x_{i+1, j} \equiv 2\left(x_{i, j}+x_{i, j+1}\right)(\bmod 3)$.
From this, we know that the number in the last row is
$$
\begin{aligned}
x_{3 n, 1} \equiv & 2^{3 n-1}\left(x_{1,1}+\mathrm{C}_{3 n-1}^{1} x_{1,2}+\cdots+\right. \\
& \left.\mathrm{C}_{3 n-1}^{3 n-2} x_{1,3 n-1}+x_{1,3 n}\right)(\bmod 3) .
\end{aligned}
$$

If $n$ is a number that meets the condition, then for any $\left(x_{1,1}, \cdots, x_{1,3 n}\right)$, it should have
$$
3 \mid x_{3 n, 1}+x_{1,1}+x_{1,3 n},
$$
$$
\text { and } x_{1,1}+x_{1,2}+\cdots+x_{1,3 n}=n \times(0+1+2)=3 n \text {. }
$$
(1) is equivalent to $3 \mid \left(2^{3 n-1}+1\right) x_{1,1}+2^{3 n-1} \mathrm{C}_{3 n-1}^{1} x_{1,2}+\cdots+$ $2^{3 n-1} \mathrm{C}_{3 n-1}^{3 n-2} x_{1,3 n-1}+\left(2^{3 n-1}+1\right) x_{1,3 n}$. Thus, combining (2), we should have
$$
\begin{array}{l}
3 \mid\left(2^{3 n-1} \mathrm{C}_{3 n-1}^{1}-2^{3 n-1}-1\right) x_{1,2}+\cdots+ \\
\left(2^{3 n-1} C_{3 n-1}^{3 n-2}-2^{3 n-1}-1\right) x_{1,3 n-1} .
\end{array}
$$

We take $\left(x_{1,1}, x_{1,2}, \cdots, x_{1,3 n}\right)=\left(0, x_{1,2}, 1, x_{1,4}, \cdots, x_{1,3 n}\right)$ and $\left(1, x_{1,2}, 0, x_{1,4}, \cdots, x_{1,3 n}\right)$, where $x_{1,2}, \cdots, x_{1,3 n}$ are arbitrary, but require that $x_{1,1}, \cdots, x_{1,3 n}$ contain exactly $n$ zeros, $n$ ones, and $n$ twos. Substituting these into (3) and then subtracting (subtracting the numerators), we get
$$
3 \mid 2^{3 n-1} \mathrm{C}_{3 n-1}^{2}-2^{3 n-1}-1
$$

Noting that $(3 n-1)(3 n-2) \equiv 2(\bmod 3) \Rightarrow C_{3 n-1}^{2}=$ $\frac{(3 n-1)(3 n-2)}{2} \equiv 1(\bmod 3)$, thus (4) becomes $3 \mid-1$, a contradiction (note that $n \geqslant 2$ is used when substituting into (3), where $x_{1,3}$ appears in the numerator).