9. (12 points) In the following equation, “迎”、“春”、“杯”、“数”、“学”、“花”、“园”、“探”、“秘” represent different non-zero digits from 1 to 9. Therefore, the maximum value of the three-digit number “迎春杯” is $\qquad$ (the “迎春杯” was founded in 1984, and this year is the 2015 “数学花园探秘”).
$$
1984-\overline{\text { 迎春杯 }}=2015-\overline{\text { 数学 }}-\overline{\text { 花园 }}-\overline{\text { 探秘 }} \text {. }
$$
【Solution】Solution: Let $\overline{\text { YCup }}=a, \overline{\text { Math }}=b, \overline{\text { Garden }}=c, \overline{\text { Mystery }}=d$,

then $a=b+c+d-31$,
to maximize $a$, the value of $b+c+d$ should be maximized, which means using the larger digits from $1 \sim 9$ to form numbers; that is, $9, 8, 7$ need to be placed in the tens place, and $6, 5, 4$ in the units place;

so the maximum value of $b+c+d$ is: $(9+8+7) \times 10+6+5+4=255$,
then $a=255-31=224$;
there are repeated digits 2 and 4, which does not meet the requirement, and the digit 3 is not used;
so the digit 3 needs to be swapped with the 4 in the units place, and 7 and 6 need to swap positions, ensuring no repeated digits. Thus, the maximum value of $b+c+d$ at this point is: $(9+8+6) \times 10+7+5+3=245$,

then $a=245-31=214$;
so the equation can be: $1984-214=2015-97-85-63$ (the number grouping is not unique)
that is, the maximum value of the three-digit number represented by "YCup" is 214.
Therefore, the answer is: 214.