8. Given that $a, b, c, d$ are all prime numbers (allowing $a, b, c, d$ to be the same), and $a b c d$ is the sum of 35 consecutive positive integers. Then the minimum value of $a+b+c+d$ is $\qquad$ .
8. 22 .
$$
\begin{array}{l}
\text { Let } a b c d=k+(k+1)+\cdots+(k+34) \\
=35(k+17),
\end{array}
$$

where $k$ is a positive integer. Then the prime numbers $a, b, c, d$ must include one 5 and one 7.
Assume $a=5, b=7$. Then $c d=k+17$.
When $k \geqslant 8$,
$$
c+d \geqslant 2 \sqrt{c d}=2 \sqrt{k+17} \geqslant 2 \sqrt{8+17}=10 \text {. }
$$

When $1 \leqslant k \leqslant 7$, there are only two cases where $k+17$ can be written as the product of two primes:
$k=4$  when, $k+17=21=3 \times 7$. In this case,
$c+d=10$;
$k=5$  when, $k+17=22=2 \times 11$. In this case,
$$
c+d=13 \text {. }
$$

In summary, the minimum value of $a+b+c+d$ is
$$
5+7+3+7=22
$$