3. Non-negative real numbers $a_{1}, a_{2}, \cdots, a_{2010}$ satisfy $\sum_{i=1}^{2010} a_{i}=1$, then the maximum value of $\sum_{i=1}^{2010} \prod_{j=0}^{1004} a_{i+j}$ is $\quad\left(a_{2010+i}=a_{i}\right)$.
3. $1005^{-1005}$.

Let $A=\sum_{i=1}^{2010} \prod_{j=0}^{1004} a_{i+j}$.
Fix $a_{2}, a_{3}, \cdots, a_{1005}, a_{1007}, \cdots, a_{2010}$, then $A=B a_{1}+C a_{1006}$, where $B, C$ are non-negative constants (determined by the fixed $a_{i}$).

It is easy to see that when one of $a_{1}, a_{1006}$ is 0, $A$ is maximized. Similarly, when one of $a_{i}, a_{1005+i} (i=2,3, \cdots, 1005)$ is 0, $A$ is maximized. At the same time, to maximize $A$, there should be 1005 consecutive terms among $a_{1}, a_{2}, \cdots, a_{2010}$ that are positive, and the other 1005 terms should be 0.

Therefore, when $a_{i}=1005^{-1}(i=1,2, \cdots, 1005)$, $a_{1005+i}=0(i=1,2, \cdots, 1005)$, $A$ reaches its maximum value $1005^{-1005}$.