15. (12 points) Given the equation $\overline{\mathrm{ABC}}-\overline{\mathrm{DE}}+\frac{\mathrm{F}^{2}}{\overline{\mathrm{GH}}}-\frac{2010}{\mathrm{I}}=12.19$ where $A, B, C, D, E, F, G, H, I$ represent distinct digits from $1 \sim 9$. Then, the five-digit number $\overline{\mathrm{ABCDE}}=$ $\qquad$
【Solution】Solution: Since the difference $12.19=12 \frac{19}{100}$, which means the difference includes $\frac{19}{100}$, the denominator after finding a common denominator is 100, meaning $G H$ is a multiple of 25. Since 2010 already contains a factor of 2, $I$ must be a multiple of 8. ($G H$ and $I$ are coprime, so they cannot be 100 and 1, 20 and 5, or 10 and 10). Therefore, $I=8, 12.19+\frac{2010}{8}=263.44$.
(1) If $G H=75, \frac{\mathrm{F}^{2}}{25}$ has a decimal part of 0.44, indicating that $F$ must be a multiple of 3, i.e., 3, 6, 9. After discussion, no such $F$ exists, so $G H=75$ does not hold.
(2) If $G H=25$, then $\frac{F^{2}}{25}$ has a decimal part of 0.44, meaning $F^2$ modulo 25 is 11, so $F=6, 263.44-\frac{36}{25}=262$.

Using the remaining $1,3,4,7,9$ to form two numbers with a difference of 262: $341-79=262$.
So the five numbers are: 34179.