19 (14 points) There are 2009 people standing in a row. Starting from the first person, they count off in a cycle of 1 to 3. Anyone who counts to 3 leaves the row, and the rest move forward to form a new row. This process continues until after the $p$-th round of counting, only 3 people remain. What were the initial positions of the last 3 people?
19 After the $p$-th round of counting, only 3 people remain, the first two of whom were originally in the first and second positions in the line.

Let the initial position of the third person be $a_{p+1}$, after the first round of counting he stands at the $a_{p}$ position, $\cdots \cdots$, after the $p$-th round of counting he stands at the $a_{1}$ position. Clearly, $a_{1}=3$.

Since $a_{p+1}, a_{p}, \cdots \cdots$ have not been eliminated, it is known that these numbers are not multiples of 3. After one round of counting, the number of positions moved from $a_{p+1}$ to $a_{p}$, $a_{p+1}-a_{p}$, is the number of multiples of 3 from 1 to $a_{p+1}$, which is $\frac{1}{3}\left(a_{p+1}-r\right)$ (where $r=1$ or 2), so $a_{p+1}-a_{p}=\frac{1}{3}\left(a_{p+1}-r\right), a_{p+1}=\frac{1}{2}\left(3 a_{p}-r\right)$ (where $r=$ 1 or 2$).

Since $a_{p+1} 、 a_{p}$ are positive integers, when $a_{p}$ is odd, $r=1$; when $a_{p}$ is even, $r=2$. Therefore,
$$
a_{p+1}=\frac{1}{2}\left(3 a_{p}-r\right) \leqslant 20092009$ until, we get $a_{2}=\frac{1}{2}(3 \times 3-1)=4, a_{3}=\frac{1}{2}(3 \times 4-2)=5, \cdots, a_{18}=$ $\frac{1}{2}(3 \times 1067-1)=1600, a_{19}=\frac{1}{2}(3 \times 1600-2)=2399>2009$.

Therefore, the last three people remaining were initially in the 1st, 2nd, and 1600th positions in the line.