Three, (20 points) The numbers $1,2, \cdots, 666$ are written on a blackboard. In the first step, the first eight numbers: $1,2, \cdots, 8$, are erased, and the sum of these numbers, 36, is written after 666; in the second step, the next eight numbers: $9,10, \cdots, 16$, are erased, and the sum of these numbers, 100, is written at the end; this process continues (i.e., in each step, the first eight numbers are erased, and the sum of these eight numbers is written at the end).
(1) How many steps are required until only one number remains on the blackboard?
(2) When only one number remains on the blackboard, find the sum of all numbers that have appeared on the blackboard (if a number appears multiple times, it should be counted repeatedly).
(1) Since each step reduces the numbers by seven, after $\frac{666-1}{7}=95$ steps, only one number remains.
(2) From $666-512=154$, it follows that after $\frac{154}{7}=22$ steps, there are 512 numbers left.

In 22 steps, a total of $22 \times 8=176$ numbers are crossed out, and their sum is
$1+2+\cdots+176=88 \times 177$.
Let $S=1+2+\cdots+666=333 \times 667$.
Then after 22 steps, the sum of the remaining 512 numbers is still $S$.

Assume there were originally $8^{k}$ numbers, with their sum being $x$. Then after $8^{k-1}$ steps, all the original $8^{k}$ numbers are crossed out, and the sum of the remaining $8^{k-1}$ numbers on the blackboard is still $x$.

Therefore, when the blackboard is left with only one number, the sum of all the numbers is $(k+1) x$.

Thus, when only one number remains on the blackboard, the sum of all the numbers that have appeared on the blackboard is
$$
\begin{array}{l}
88 \times 177+4 S=88 \times 177+4 \times 333 \times 667 \\
=904020
\end{array}
$$