2. Consider the binomial coefficients $\mathrm{C}_{n}^{0}, \mathrm{C}_{n}^{1}, \cdots, \mathrm{C}_{n}^{n}$ as a sequence. When $n \leqslant 2004\left(n \in \mathbf{N}_{+}\right)$, the number of sequences in which all terms are odd is $\qquad$ groups.
2.10.

From $\mathrm{C}_{n}^{r}=\mathrm{C}_{n-1}^{r-1}+C_{n-1}^{r}$, we know that the 2004 numbers $n=1,2, \cdots, 2004$ can be arranged in the shape of Pascal's Triangle, and the even and odd numbers can be represented by 0 and 1, respectively, as shown in Figure 3.
The pattern of the changes in the parity of each term is:
When and only when $n=2^{k}-1\left(k \in \mathbf{N}_{+}\right)$, all terms are odd. Solving $2^{k}-1 \leqslant 2004$, we get
$$
k=1,2,3,4,5,6,7,8,9,10 .
$$

Therefore, there are 10 sequences where all terms are odd.