2. In the sequence $\left\{a_{n}\right\}$,
$$
a_{1}=\frac{1}{3}, a_{n+1}=2 a_{n}-\left[a_{n}\right] \text {, }
$$

where, $[x]$ denotes the greatest integer not exceeding the real number $x$. Then
$$
a_{2009}+a_{2010}=
$$
2. 2009.

Given $a_{1}=\frac{1}{3}, a_{2}=\frac{2}{3}, a_{3}=\frac{4}{3}$.
We will prove by mathematical induction:
$$
a_{n+2}-a_{n}=1, a_{n}+a_{n+1}=n \text {. }
$$

Obviously, when $n=1$, the conclusion holds.
Assume when $n=k$, the conclusion holds, i.e.,
$$
a_{k+2}-a_{k}=1, a_{k}+a_{k+1}=k \text {. }
$$

Then when $n=k+1$,
$$
\begin{array}{l}
a_{k+3}-a_{k+1} \\
=2 a_{k+2}-\left[a_{k+2}\right]-\left(2 a_{k}-\left[a_{k}\right]\right) \\
=2\left(a_{k+2}-a_{k}\right)-\left(\left[a_{k+2}\right]-\left[a_{k}\right]\right) \\
=2-\left(\left[a_{k}+1\right]-\left[a_{k}\right]\right)=1, \\
a_{k+1}+a_{k+2}=a_{k+1}+\left(a_{k}+1\right)=k+1 .
\end{array}
$$

Thus, when $n=k+1$, the conclusion also holds.
In summary, $a_{n+2}-a_{n}=1, a_{n}+a_{n+1}=n$ always holds. Therefore, $a_{2009}+a_{2010}=2009$.