6. Fill the 9 squares in the vertical addition below with the digits 1 to 9, using each digit only once. The maximum possible sum is $\qquad$


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【Analysis】The maximum hundred's place of the sum is 9.
Considering the hundred's place of the three-digit addend, it can be 7 or 8,
If it is 7, then the sum will not exceed $82+705+3164=3951$, meaning the maximum ten's place of the sum is 5.
If it is 8, then the ten's place of the sum can be 7.
The remaining digits are 1, 2, 4, 5, 6,
Considering the unit's place of the sum, if the unit's place is 6, two digits need to sum to 4 or 14, which is clearly impossible.
If the unit's place is 5, two digits need to sum to 3 or 13, only $1, 2$ can, but the ten's place cannot be satisfied.
If the unit's place is 4, two digits need to sum to 2 or 12, which is clearly impossible.
If the unit's place is 2, two digits need to sum to 0 or 10, only $4, 6$ can, at this point the ten's place is $0, 1, 5$, plus the carry-over 1 from the unit's place, it is exactly 7.
Thus, the maximum sum is $12+804+3156=3972$.