6.2. In a $2 \times 2$ grid, 4 positive integers are filled in, where the difference between any two numbers in the same column is 6, and the difference between any two numbers in the same row is 1 times (i.e., one number is twice the other). Try to determine: which numbers are filled in the grid? Can there be different sets of numbers?
6.2, there is only one way to fill:

In the grid, there are only 6 and 12, as shown in Figure 2(a).
\begin{tabular}{|c|c|}
\hline 6 & 12 \\
\hline 12 & 6 \\
\hline
\end{tabular}
(a)
\begin{tabular}{|c|c|}
\hline$x$ & $2 x$ \\
\hline$x \pm 6$ & $2 x \pm 6$ \\
\hline
\end{tabular}
(b)

Figure 2
The following proves that there are no other ways to fill. Suppose the first row is filled with $x$ and $2 x$ (in Figure 2(b), with $x$ on the left and $2 x$ on the right as an example, the opposite case is similar), then below $x$, only $x+6$ or $x-6$ can be filled, and below $2 x$, only $2 x+6$ or $2 x-6$ can be filled. Since in the second row, one number is twice the other, it can only be one of the following 8 cases:
$$
\begin{array}{l}
x+6=2(2 x+6), x-6=2(2 x+6), \\
x+6=2(2 x-6), x-6=2(2 x-6), \\
2(x+6)=2 x+6,2(x-6)=2 x+6, \\
2(x+6)=2 x-6,2(x-6)=2 x-6 .
\end{array}
$$

Among them, only one equation has a positive integer solution, and the other cases do not meet the requirements.