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 ( ).
【Answer】1798.
【Analysis】Key point: Number Puzzles, Maximum and Minimum Problems
Corresponding Lecture in Xueersi: Number Puzzles in Grade 3 Summer Class (Lecture 8), Calculation in Short-term Competition Class (Lecture 1)
The sum of the digits of the three addends is $1+2+3+\ldots+9=45$, and the digit sum of their sum 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 digits of the addends cannot equal 1, so there must be a carry-over of 1 from the thousands place, making the thousands digit of the four-digit number 1.
(2) If there is a carry-over of 2 at the tens place, then the sum of the three digits at 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 at the hundreds place + carry-over $2=11$, the hundreds digit of the three-digit number is at least 2, so the hundreds digit 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 at the tens place + carry-over $1=21$, so the tens digit 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 digit of the four-digit number can be at most $8(8+3+4=15$, carrying 1 to the tens place).