7. (10 points) Five football teams compete, with each pair of teams playing one match: the winner of each match gets 3 points, the loser gets 0 points, and in the case of a draw, both teams get 1 point. After all the matches are completed, it is found that the five teams' points are five consecutive natural numbers. Let the 1st, 2nd, 3rd, 4th, and 5th place teams have drawn $A$, $B$, $C$, $D$, and $E$ matches respectively. Then the five-digit number $\overline{\mathrm{ABCDE}}=$ $\qquad$
【Solution】Solution: Five football teams compete, for a total of $C_{5}^{3}=10$ matches. The sum of points for each match is 2 or 3, so the total points are between 20 and 30.

The points of the five teams are exactly five consecutive natural numbers, and the sum of the points of the five teams is 20 to 30, with the following three scenarios: 2 to 6, 3 to 7, 4 to 8.

If the points of the five teams are $2 \sim 6$, then the total points are 20, which means all matches are draws, and each team gets 4 points, a contradiction. If the points of the five teams are $4 \sim 8$, then the total points are 30, which means there are no draws, and each team's points are multiples of 3, a contradiction. Therefore, the points of the five teams are $3 \sim 7$, with a total of 25 points, 5 draws, and $A+B+C+D+E=2 \times 5=10$. The first place gets 7 points, with 4 matches, which can only be 2 wins, 1 draw, and 1 loss, so $A=1$,

The third place gets 5 points, with 4 matches, which can only be 1 win, 2 draws, and 1 loss, so $C=2$, The fourth place gets 4 points, with 4 matches, if it is 4 draws, then $B \geqslant 3, D=4, E=3$, so $A+B+C+D+E \geqslant 1+3+2+4+1 > 10$, a contradiction, so the fourth place is 1 win, 1 draw, and 2 losses, thus $D=1$, so $B+E=10-A-C-D=6$, and $B \leqslant 3, E \leqslant 3$, which can only be $B=E=3$, In summary, the five-digit number $\overline{\mathrm{ABCDE}}=13213$.