For a positive integer $ n$, let $ \theta(n)$ denote the number of integers $ 0 \leq x < 2010$ such that $ x^2 \minus{} n$ is divisible by $ 2010$. Determine the remainder when $ \displaystyle \sum_{n \equal{} 0}^{2009} n \cdot \theta(n)$ is divided by $ 2010$.

Let's think step by step and output the final answer within \boxed{}.
user