Solution:

Answer: $335$

Let us consider the sum $\sum_{n=0}^{2009} n \cdot \theta(n) \pmod{2010}$ in another way. Consider the sum $0^{2} + 1^{2} + 2^{2} + \cdots + 2009^{2} \pmod{2010}$. For each $0 \leq n < 2010$, in the latter sum, the term $n$ appears $\theta(n)$ times, so the sum is congruent to $\sum_{n=0}^{2009} n \cdot \theta(n)$. In other words,
$$\sum_{n=0}^{2009} n \cdot \theta(n) = \sum_{n=0}^{2009} n^{2} = \frac{(2009)(2009+1)(2 \cdot 2009+1)}{6} \equiv (-1) \cdot \frac{2010}{6} \cdot (-1) = 335 \pmod{2010}$$