6. The ball invited 2018 couples, who were arranged in areas numbered $1,2, \cdots, 2018$. The ball organizers stipulated: when the ball reaches the $i$-th minute, the couple standing in area $s_{i}$ (if there is someone in this area) will move to area $r_{i}$, and the couple originally in area $r_{i}$ (if there is someone in this area) will exit the dance, where,
$$
s_{i} \equiv i(\bmod 2018), r_{i} \equiv 2 i(\bmod 2018),
$$

and $1 \leqslant s_{i} 、 r_{i} \leqslant 2018$. According to this rule, when the ball has been going on for $2018^{2}$ minutes, how many couples are still dancing (Note: if $s_{i}=r_{i}$, then the couple in area $s_{i}$ remains in the original area and does not exit the dance)?
6. When the ball has been going on for $2018^{2}$ minutes, there are still 505 couples dancing.

According to the problem, at the 2018th minute of the ball, the remaining couples will be in areas $2,4, \cdots, 2018$; at the 3027th minute, the remaining couples will be in areas $1010,1012, \cdots, 2018$; at the 4036th minute, the remaining couples will be in areas $2,6, \cdots, 2018$.

Next, prove that no more couples will drop out of the dance in the following 1009 minutes.

In fact, at the 4036th minute, the 505 remaining couples are in areas numbered $2 \times 1,2 \times 3, \cdots$, $2 \times 1009$, and the largest odd factor of the area numbers they are in are all different. From the 4036th minute onwards for 1009