【Answer】 8543
【Analysis】Test point: Number puzzle
1. From the hundred's place without carrying, then “Zhong > Bei”
2. From “Zhong > Bei”, then “Zhong + Huan” carries over
3. 1) When “$7 +$ Zhong” does not carry, “$7 +$ Zhong = Huan”, by “Zhong + Huan $=$ Bei +10”, then “Zhong $+7+$ Zhong $=$ Bei +10”
Therefore: $2 \times$ Zhong $=$ Bei +3
When Bei $=1$, exclude
When Bei $=3$, then Zhong $=3$, exclude
When Bei $=5$, then Zhong $=4$, Huan $=1$, exclude
When Bei $=7$, exclude
When Bei $=9$, then Bang $=0$, exclude
2) When “$7 +$ Zhong” carries over, “$7 +$ Zhong $=$ Huan +10”, by “Zhong + Huan $+1=$ Bei +10”, then “Huan $+3+$ Huan $+1=$ Bei +10”,
Therefore $2 \times$ Huan $=$ Bei +6
When Bei $=2$, then Huan $=4$, Zhong $=7$, exclude
When Bei $=4$, then Huan $=5$, Zhong $=8$, Bang $=3$, correct
When Bei $=6$, then Huan $=6$, exclude
When Bei $=8$, then Huan $=7$, exclude
Therefore, “Zhong Huan Bei Bang”=8543
14. As shown in the figure, the same Chinese character represents the same digit, and different Chinese characters represent different digits. All Chinese characters do not represent 0, nor do they represent the digits that have already appeared in the figure. Then the four-digit number “中环杯棒”= $\qquad$

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