Step 1: Turn each tile into a single number.
For every 3 by 3 tile, just count how many of its nine dots are filled in black. This turns Panel I into a 3 by 3 table of numbers instead of pictures.
\[ \begin{matrix} 8 & 1 & 6 \\ 3 & 5 & x \\ 4 & 9 & y \end{matrix} \]
Here $x$ is the top missing tile and $y$ is the bottom missing tile.
Step 2: Use two known columns to find the target total.
The first column reads $8, 3, 4$ and adds to $15$. The second column reads $1, 5, 9$ and also adds to $15$. Since two different columns both land on the same total, the whole table must be built so that every line, whether a row or a column, adds up to $15$.
Step 3: Solve for the missing tile counts using the third column and the bottom row.
The third column gives $6+x+y=15$, so $x+y=9$. The bottom row gives $4+9+y=15$, so $y=2$. Putting $y=2$ back into $x+y=9$ gives $x=7$.
Step 4: Cross check with the anti diagonal.
The anti diagonal running through $6, 5, 4$ adds to $15$ as well, which backs up the rule used above and shows the whole table behaves like a balanced 3 by 3 number square with every line equal to $15$.
Step 5: Compare with the four choices.
We need a top tile with $7$ filled dots sitting above a bottom tile with $2$ filled dots. Checking the images, option (i) is the only pair with exactly $7$ dots on top and $2$ dots below it. The other three options give either the wrong top count or the wrong bottom count.
\[ \boxed{\text{(i)}} \]
