Let us consider the $\operatorname{sum} \sum_{n=0}^{2009} n \cdot \theta(n)(\bmod 2010)$ in a another way. Consider the sum $0^{2}+1^{2}+2^{2}+\cdots+2007^{2}(\bmod 2010)$. 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, $$\sum_{n=0}^{2009} n \cdot \theta(n)=\sum_{n=0}^{2009} n^{2}=\frac{(2009)(2009+1)(2 \cdot 2009+1)}{6} \equiv(-1) \cdot \frac{2010}{6} \cdot(-1)=335 \quad(\bmod 2010)$$