Example 3 Given that $a, b, c, d$ are all prime numbers (allowing $a, b, c, d$ to be the same), and $abcd$ is the sum of 35 consecutive positive integers. Then the minimum value of $a+b+c+d$ is $\qquad$. ${ }^{[3]}$
(2011, Xin Zhi Cup Shanghai Junior High School Mathematics Competition)
【Analysis】According to the problem, we set
$$
\begin{array}{l}
a b c d=k+(k+1)+\cdots+(k+34)\left(k \in \mathbf{N}_{+}\right) \\
\Rightarrow \frac{(2 k+34) \times 35}{2}=a b c d \\
\Rightarrow(k+17) \times 5 \times 7=a b c d .
\end{array}
$$

By symmetry, without loss of generality, let $c=5, d=7, a \leqslant b$.
We only need to find $(a+b)_{\text {min }}$.
Also, $a+b \geqslant 2 \sqrt{a b}=2 \sqrt{k+17}$
$\geqslant 2 \sqrt{18}>2 \sqrt{16}=8$.
Thus, $a+b>8$, i.e., $a+b \geqslant 9$.
If $a+b=9$, then $a=2, b=7$.
This contradicts $a b=k+17>17$.
Therefore, $a+b \geqslant 10$.
When $k=4$, $a=3, b=7$, the equality holds.
Thus, $(a+b+c+d)_{\min }=22$.