Eight, (25 points) Find the smallest positive integer that is divisible by 209 and whose sum of digits equals 209.
(Zhang Lei provided the question)
Eight, the smallest number is $2 \times 10^{24} + 2 \times 10^{23} - 10^{15} - 1$. Prove: Since
$$
209 = 11 \times 19, 209 = 9 \times 23 + 2 \text {, }
$$

the number is at least 24 digits long and is divisible by 11 and 19.
(1) If the number is a 24-digit number, let the $i$-th digit from right to left be $a_{i} (1 \leqslant i \leqslant 24)$, and the number be $S$. Then
$$
\begin{array}{l}
S = \sum_{i=1}^{24} 10^{i-1} a_{i} \equiv \sum_{i=1}^{24}(-1)^{i-1} a_{i} \\
\equiv 0(\bmod 11) .
\end{array}
$$

Let $S_{1} = a_{1} + a_{3} + \cdots + a_{23}$,
$$
S_{2} = a_{2} + a_{4} + \cdots + a_{24} \text {. }
$$

Then $S_{1} \equiv S_{2}(\bmod 11)$.
Since $S_{1} + S_{2} = 209$, and the maximum of $S_{1}$ and $S_{2}$ is no more than 108, the minimum is no less than 101, and the absolute difference is no more than 7. Since $S_{1}$ and $S_{2}$ are one odd and one even, $S_{1} - S_{2} \neq 0$, i.e., $S_{1} \neq \equiv S_{2}(\bmod 11)$, which is a contradiction.
Therefore, the number that satisfies the condition must be at least 25 digits long.
(2) If the number is a 25-digit number, similarly, let the number be $S$,
$$
\begin{array}{l}
S_{1} = a_{1} + a_{3} + \cdots + a_{25}, \\
S_{2} = a_{2} + a_{4} + \cdots + a_{24} .
\end{array}
$$
1) If $a_{25} = 1$, since the maximum of $S_{1}$ and $S_{2}$ is no more than 109, the minimum is no less than 100, and the absolute difference is no more than 9. Since $S_{1}$ and $S_{2}$ are one odd and one even, $S_{1} - S_{2} \neq 0$, i.e., $S_{1} \neq \equiv S_{2}(\bmod 11)$. In this case, there is no number that satisfies the condition.
2) If $a_{25} = 2$, since the maximum of $S_{1}$ and $S_{2}$ is no more than 110, the minimum is no less than 99, and the absolute difference is no more than 11. Since $S_{1}$ and $S_{2}$ are one odd and one even, $S_{1} - S_{2} \neq 0$, only $S_{1} = 110, S_{2} = 99$ might satisfy the condition. In this case, $a_{1} = a_{3} = \cdots = a_{23} = 9$.
(i) If $a_{24} = 0$, then the number is
$$
S = 2 \times 10^{24} + 10^{23} - 1 \text {, }
$$

which leaves a remainder of 5 when divided by 19, and does not satisfy the condition.
(ii) If $a_{24} = 1$, then the number is
$$
S = 2 \times 10^{24} + 2 \times 10^{23} - 1 - 10^{x} \text {, }
$$

where $x$ is odd.
Since $2 \times 10^{24} + 2 \times 10^{23} - 1 \equiv 8(\bmod 19)$, and the remainder of $10^{k}$ modulo 19 is $10, 5, 12, 6, 3, 11, 15, 17, 18, 9, 14, 7, 13, 16, 8, 4, 2, 1$ in a cycle, thus, $x = 18t + 15$. Hence, $x = 15$. In this case, the number that satisfies the condition is $2 \times 10^{24} + 2 \times 10^{23} - 10^{15} - 1$.
In summary, the smallest number that satisfies the condition is
$$
2 \times 10^{24} + 2 \times 10^{23} - 10^{15} - 1 \text {. }
$$
(Zhang Tongjun provided)