1. **Understanding the function $b(p)$**:
- The function $b(p)$ is defined such that $b(p) = k$ where $k$ is the unique positive integer satisfying $|k - \sqrt{p}| < \frac{1}{2}$.
- This implies $k - \frac{1}{2} < \sqrt{p} < k + \frac{1}{2}$.
- Squaring all parts of the inequality, we get:
$$(k - \frac{1}{2})^2 < p < (k + \frac{1}{2})^2$$
- Simplifying the squared terms:
$$k^2 - k + \frac{1}{4} < p < k^2 + k + \frac{1}{4}$$
- Since $p$ is an integer, we can ignore the fractional parts:
$$k^2 - k + 1 \leq p \leq k^2 + k$$

2. **Counting the number of $p$ values for each $k$**:
- For each $k$, the number of integers $p$ in the range $k^2 - k + 1 \leq p \leq k^2 + k$ is:
$$(k^2 + k) - (k^2 - k + 1) + 1 = 2k$$
- Therefore, there are $2k$ values of $p$ for which $b(p) = k$.

3. **Summing $b(p)$ for $p$ from 1 to 2007**:
- We need to find the sum $S = \sum_{p=1}^{2007} b(p)$.
- We observe that $b(p) = k$ for $2k$ values of $p$.
- We need to determine the largest $k$ such that $k^2 + k \leq 2007$:
$$k^2 + k - 2007 \leq 0$$
- Solving the quadratic equation $k^2 + k - 2007 = 0$ using the quadratic formula:
$$k = \frac{-1 \pm \sqrt{1 + 4 \cdot 2007}}{2} = \frac{-1 \pm \sqrt{8029}}{2}$$
- Approximating $\sqrt{8029} \approx 89.6$:
$$k \approx \frac{-1 + 89.6}{2} \approx 44.3$$
- Thus, the largest integer $k$ is 44.

4. **Calculating the sum $S$**:
- For $k = 1$ to $44$, each $k$ contributes $2k \cdot k = 2k^2$ to the sum.
- The sum of squares of the first 44 integers is:
$$\sum_{k=1}^{44} 2k^2 = 2 \sum_{k=1}^{44} k^2 = 2 \cdot \frac{44 \cdot 45 \cdot 89}{6} = 2 \cdot 29370 = 58740$$
- For $p$ from 1981 to 2007 (27 values), $b(p) = 45$:
$$27 \cdot 45 = 1215$$
- Adding these contributions:
$$S = 58740 + 1215 = 59955$$

5. **Finding the remainder when $S$ is divided by 1000**:
- Calculating $59955 \mod 1000$:
$$59955 \equiv 955 \pmod{1000}$$

The final answer is $\boxed{955}$.