13. Fill in the following squares with $1,2,3,4,5,6,7,8,9$, using each number only once, then the largest four-digit number is ( ).
口口口口+口口口+口口=2115

Note: The squares (口) represent blank spaces to be filled with the numbers.
【Answer】1798.
【Analysis】Exam point: Number puzzle, Maximum value problem
The sum of the digits of the three addends is $1+2+3+\ldots+9=45$, and the sum of the digits of their total 2115 is $2+1+1+5=9$.
The three addends can only produce carry-overs at the tens, hundreds, and thousands places, and the sum of these carry-overs is $(45-9) \div 9=4$.
(1) Since the hundreds place of 2115 is 1, the sum of the hundreds place digits of the addends cannot equal 1, so there must be a carry-over of 1 from the thousands place, making the thousands place of the four-digit number 1.
(2) If there is a carry-over of 2 from the tens place, then the sum of the three digits in the units place must be 25, but the largest three digits $9+8+7<25$, so the tens place can only carry over 1, and the hundreds place must carry over 2. Then, the sum of the two digits in the hundreds place + the carry-over $2=11$, the hundreds place of the three-digit number is at least 2, so the hundreds place of the four-digit number is at most $7(7+2+$ carry-over $2=11$, carrying 1 to the thousands place).
(3) The sum of the three digits in the tens place + the carry-over $1=21$, so the tens place of the four-digit number can be at most 9 (9+ carry-over $1+6+5=21$, carrying 2 to the hundreds place).
(4) The units place of the four-digit number can be at most $8(8+3+4=15$, carrying 1 to the tens place).