7. Given that the 6027-digit number $\frac{a b c a b c \cdots a b c}{2000 \uparrow a b c}$ is a multiple of 91. Find the sum of the minimum and maximum values of the three-digit number $\overline{a b c}$.
7. From $91=7 \times 13, 1001=7 \times 11 \times 13$, we know 91 | 1001.
And $\overline{a b c a b c}=1001 \times \overline{a b c}$, thus, $91 \mid \overline{a b c a b c}$. $2009 \uparrow a b c$
$$
\overline{a_{2009 \uparrow a b c}^{a b c a b c}}=\underset{2 \times 1004 \uparrow a b c}{\overline{a b c a b c \cdots a b c}} \times 1000+\overline{a b c},
$$

Since $\overline{a b c a b c \cdots a b c}$ is divisible by 91, then $\overline{a b c}$ is divisible by 91.
$2 \times 1004 \uparrow a b$.
The smallest three-digit number divisible by 91 is 182, and the largest is 910, so the sum of the minimum and maximum values of $\overline{a b c}$ is 1092.