8. If the remainder of $\underbrace{11 \cdots 1}_{n+1 \uparrow} 1$ divided by 3102 is 1, then the smallest positive integer $n$ is $\qquad$ .
8. 138 .

Notice that, $3102=2 \times 3 \times 11 \times 47$.
From $\underbrace{11 \cdots 1}_{n+1 \uparrow}=3102 k+1(k \in \mathbf{Z})$, we know $\underbrace{11 \cdots 10}_{n \uparrow}=3102 k$.
Thus, $\underbrace{11 \cdots 10}_{n \uparrow}$ is divisible by $2, 3, 11, 47$.
(1) For any positive integer $n$, obviously, $\underbrace{11 \cdots 10}_{n \uparrow}$ is divisible by 2;
(2) $\underbrace{11 \cdots 10}_{n \uparrow}$ is divisible by 3 if and only if $31 n$;
(3) $\underbrace{11 \cdots 10}_{n \uparrow}$ is divisible by 11 if and only if $2 \mid n$;
(4) Also, $\underbrace{11 \cdots 10}_{n \uparrow}=\frac{1}{9}\left(10^{n+1}-10\right)$,
$$
(9,47)=1,(10,47)=1 \text {, }
$$

then $47|\underbrace{11 \cdots 10}_{n \uparrow} \Leftrightarrow 47|\left(10^{n}-1\right)$.
By Fermat's Little Theorem, $10^{46} \equiv 1(\bmod 47)$.
Let $t$ be the smallest positive integer such that $10^{t} \equiv 1(\bmod 47)$. Then $t \mid 46$.
And $10 \equiv 10(\bmod 47)$,
$10^{2} \equiv 6(\bmod 47)$,
$10^{23} \equiv 46(\bmod 47)$,
thus $t=46$.
Therefore, $46|n \Leftrightarrow 47| \underbrace{11 \cdots 1}_{n \uparrow}$.
In summary, $n_{\min }=[2,3,46]=138$.