1. Let the times for the particle to move to $A_{1}, A_{2}, \cdots, A_{n}$ be $a_{1}, a_{2}, \cdots, a_{n}$, respectively, then $a_{1}=2, a_{2}=6, a_{3}=12, a_{4}=$ $20, \cdots$, and $a_{n}-a_{n-1}=2 n$. Therefore, we get the recurrence relation
$$
\left\{\begin{array}{l}
a_{n}=a_{n-1}+2 n \\
a_{1}=2
\end{array}\right.
$$

So $a_{n}=a_{1}+\left(a_{2}-a_{1}\right)+\left(a_{3}-a_{2}\right)+\cdots+\left(a_{n}-a_{n-1}\right)=2+2 \times(2+3+4+\cdots+n)$,
which means $a_{n}=n \times(n+1)$.
Since $44 \times 45=1980$, the particle reaches point $A_{44}(44,44)$ after 1980 seconds. According to the movement pattern, at points $A_{1}, A_{2}, \cdots, A_{n}$, it moves downward at odd points and to the left at even points. Therefore, upon reaching point $A_{44}(44,44)$, it moves left for 24 seconds to reach point $(20, 44)$, which means it has moved for 2004 seconds. Thus, the particle's position is at point $(20,44)$.
1. As shown in the figure, a particle moves in the first quadrant. In the first second, it moves from the origin to $(0,1)$, and then it continues to move back and forth in directions parallel to the $x$-axis and $y$-axis, i.e., $(0,0) \rightarrow(0,1) \rightarrow(1,1) \rightarrow(1,0) \rightarrow(2,0) \rightarrow \cdots)$, and it moves one unit length per second. Therefore, at 2004 seconds, the position of this particle is $\qquad$.