Solution:
The key idea is to pair up the terms $\left\lfloor \frac{2025}{x} \right\rfloor$ and $\left\lfloor \frac{2025}{x} \right\rfloor$. There are 1000 such pairs and one lone term, $\left\lfloor \frac{2025}{1000.5} \right\rfloor = 2$. Thus,
$$\sum_{j = -1000}^{1000} \left\lfloor \frac{2025}{j + 0.5} \right\rfloor = 2 + \sum_{x \in \{0.5,1.5, \ldots , 999.5\}} \left(\left\lfloor \frac{2025}{x} \right\rfloor + \left\lfloor \frac{2025}{-x} \right\rfloor \right).$$
We note that
Therefore,
As $x$ ranges in the set $\{0.5, 1.5, 2.5, \ldots , 999.5\}$, $2x$ ranges in the set $\{1, 3, 5, \ldots , 1999\}$. This set includes all 15 odd divisors of 4050 except for 2025. Thus, there are 14 values of $x$ for which $\left\lfloor \frac{2025}{x} \right\rfloor + \left\lfloor \frac{2025}{- x} \right\rfloor$ evaluates to 0, and the remaining $1000 - 14 = 986$ values of $x$ make it evaluate to $-1$. Therefore,
$$\sum_{j = -1000}^{1000} \left\lfloor \frac{2025}{j + 0.5} \right\rfloor = 2 + \sum_{x \in \{0.5,1.5, \ldots , 999.5\}} \left(\left\lfloor \frac{2025}{x} \right\rfloor + \left\lfloor \frac{2025}{-x} \right\rfloor \right) = 2 + 986 \cdot (-1) = \boxed{-984}.$$