9. (12 points) Select 8 different digits from the numbers $1-9$ to fill in the squares below, so that the vertical equation holds. The largest possible four-digit number is $\qquad$ .



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【Solution】Solution: First, analyze the thousands digit of the four-digit number is 1, at this point there are still 8 digits from 2 to 9. Next, look at the sum of the last three digits of the individual positions, which ends in 0. We can find $(2,3,5),(3,8,9),(4,7,9),(5,6,9)$, and $(5,7,8)$, a total of 5 combinations.

Next, the tens digit of 2010 is 1, which means there must be a carry-over. In the result, the hundreds digit is 0, so the hundreds digit of the four-digit number can be at most 7, and the hundreds digit of the three-digit number can be 2. Considering the tens digit of 2010 is 1, and there is a carry-over of 2 from the units, the sum of the two tens digits must be 9. The tens digit of the four-digit number can be at most 6. The digits used so far are $1,2,7,3,6$, and the largest possible units digit is a combination of $4,7,9$, with the units digit being 9.

Thus: The largest four-digit number is 1769.
Therefore, the answer is: 1769