Assuming the sets are distinguishable; otherwise, divide the answer by $6$ (define the first set as the one containing $1$ and the second set as the one containing $2$). $1$ has $3$ choices to go. After this, every integer $n \ge 2$ has $2$ choices to go, namely the two sets that $n-1$ is not in. So there are $3 \cdot 2^{2005}$ ways. However, some of these methods give some empty set. We will count the number of ways that give at least one empty set. First, if there are at least two empty sets, it's impossible; the only remaining set must contain all numbers, which means it contains $1,2$. Now we will compute the number of ways where there is one empty set. There are $3$ choices to choose the empty set. After this, $1$ has $2$ choices, namely the two sets that are declared non-empty. After this, every integer $n \ge 2$ only has one choice to go; namely, the empty set that $n-1$ is not contained in. So there are $6$ ways where there is at least one empty set. So in total there are $3 \cdot 2^{2005} - 6$ ways. (Or simplifying, $6 (2^{2004} - 1)$, to make it more obvious that it's divisible by $6$. Your choice.)