1. There are $n$ cards on the table, each written with a positive integer from $1 \sim n$, and the numbers on each card are distinct. Initially, the cards are stacked in a pile, with the numbers on the cards arranged in increasing order from top to bottom. Operation $A$ is to move the top card of the pile to the bottom of the pile, and operation $B$ is to remove the top card from the pile. Repeatedly perform the operations $A B B A B B A B B A B B \cdots$, until only one card remains on the table. Let $L(n)$ be the number on the last remaining card. Find all positive integers $k$ such that
$$
L(3 k)=k \text {. }
$$
1. Consider the positive integer $n (n \geqslant 3)$ that satisfies $L(n)=1$. After each card in the stack has been operated on once, the card with the number 1 should still be the top card. Therefore, $n$ must be a multiple of 3.

Thus, the remaining number of cards $\frac{n}{3}$ should be a multiple of 1, 2, or 3. This indicates that the necessary and sufficient condition for a positive integer $n$ to satisfy $L(n)=1$ is $n=3^{j}$ or $2 \times 3^{j}$ (where $j$ is any non-negative integer).
Assume the positive integer $k$ satisfies $L(3 k)=k$.
Consider the moment after $k-1$ operations, at this point, the top card in the stack on the table has the number $k$.

Since this card must be left, $k-1$ must be a multiple of 3, and operation $A$ has been performed $\frac{k-1}{3}$ times before this, with $2 k+1$ cards still not operated on, bearing the numbers $k, k+1, \cdots, 3 k$.
Thus, there are
$$
\frac{k-1}{3}+(2 k+1)=\frac{7 k+2}{3}
$$

cards remaining.
Subsequently, the operations are repeated until the top card is left, then
$$
L\left(\frac{7 k+2}{3}\right)=1 .
$$

For non-negative integers $j$,
$$
\begin{array}{l}
\frac{7 k+2}{3}=3^{j} \text { or } 2 \times 3^{j} \\
\Rightarrow k=\frac{3^{j+1}-2}{7} \text { or } \frac{2 \times 3^{j+1}-2}{7} .
\end{array}
$$

Since $k$ is a positive integer, $3^{j+1}-2$ and $2 \times 3^{j+1}-2$ should be divisible by 7.

Therefore, for all non-negative integers $j, k=\frac{3^{6 j+2}-2}{7}$ or for all positive integers $j, k=\frac{2 \times 3^{6 j}-2}{7}$.