2. Let the set $M=\left\{x \mid x=2^{n}-2^{k}\right.$, where $n, k \in$ $\mathbf{N}$, and $n>k\}, P=\{x \mid 1912 \leqslant x \leqslant 2004$, and $x$ $\in \mathbf{N}\}$. Then, the sum of all elements in the set $M \cap P$ is $\qquad$.
2.3904.

Since $2^{10}=1024,2^{11}=2048$, all elements in the set $M \cap P$ must be of the form $2^{11}-2^{k}$ (where $k \in \mathbf{N}$).
Also, $1912 \leqslant 2^{11}-2^{k} \leqslant 2004$, so $44 \leqslant 2^{k} \leqslant 136$.
The only values of $k$ that satisfy this condition are 6 and 7, so the set $M \cap$ $P$ contains only two elements, which are
$$
2^{11}-2^{6}=1984 \text { and } 2^{11}-2^{7}=1920 \text {. }
$$