Answer: B

Small Hint:

If $a_k$ summands have $7$ in decimal place $k$, divide the sum by $7$

Big Hint:

Compare the resulting digit counts with the repeating decimal for $\frac{1}{7}$, then seek a six-digit repeating construction

Solution:

Suppose $1$ is a sum of $n$ special numbers, and let $a_k$ count summands having a $7$ in the $k$th decimal place. Dividing by $7$ gives $$\begin{aligned} \frac17&=\frac{a_1}{10}+\frac{a_2}{10^2}+\cdots\\ &=0.\overline{142857}. \end{aligned}$$ For $n\le9,$ each $a_k$ is a digit, so $a_1,a_2,\ldots=1,4,2,8,5,7,\ldots;$ hence $n\ge8.$ Eight suffice because the repeating special decimals represented by $$\begin{aligned} 700700+2(070707)\\ {}+2(077777)\\ {}+3(000777)&=999999. \end{aligned}$$ Their sum is $1.$ Therefore the minimum is $8,$ and B is correct.