Let $n$ be positive numbers. Consider the following region on $xy$ plane.

\[ D_n=\left\{(x,y) |\ \frac{x}{n+\frac{1}{2}}\leq y\leq [x+1]-x,\ x\geq 0 \right\}. \]

Here $\ [x]$ is the greatest integer $\ \leq x.$ Find the area of $D_n.$
1. **Understanding the Region \( D_n \)**:
   The region \( D_n \) is defined by the set of points \((x, y)\) in the \(xy\)-plane that satisfy the following conditions:
   \[
   \frac{x}{n+\frac{1}{2}} \leq y \leq \lfloor x+1 \rfloor - x \quad \text{and} \quad x \geq 0
   \]
   Here, \(\lfloor x \rfloor\) denotes the greatest integer less than or equal to \(x\).

2. **Analyzing the Upper Bound**:
   The upper bound of \(y\) is given by:
   \[
   y \leq \lfloor x+1 \rfloor - x
   \]
   Since \(\lfloor x+1 \rfloor\) is the greatest integer less than or equal to \(x+1\), it can be written as:
   \[
   \lfloor x+1 \rfloor = \lfloor x \rfloor + 1
   \]
   Therefore, the upper bound becomes:
   \[
   y \leq (\lfloor x \rfloor + 1) - x = 1 + \lfloor x \rfloor - x
   \]

3. **Analyzing the Lower Bound**:
   The lower bound of \(y\) is given by:
   \[
   y \geq \frac{x}{n+\frac{1}{2}}
   \]

4. **Finding the Area**:
   To find the area of \(D_n\), we need to integrate the difference between the upper and lower bounds of \(y\) over the range of \(x\).

5. **Setting Up the Integral**:
   The region \(D_n\) can be divided into intervals based on the integer values of \(x\). For each integer \(k\), where \(k \leq x < k+1\), the bounds of \(y\) are:
   \[
   \frac{x}{n+\frac{1}{2}} \leq y \leq 1 + k - x
   \]
   The area for each interval \([k, k+1)\) can be found by integrating the difference between the upper and lower bounds of \(y\):
   \[
   \text{Area} = \int_k^{k+1} \left( (1 + k - x) - \frac{x}{n+\frac{1}{2}} \right) \, dx
   \]

6. **Evaluating the Integral**:
   \[
   \text{Area} = \int_k^{k+1} \left( 1 + k - x - \frac{x}{n+\frac{1}{2}} \right) \, dx
   \]
   Simplifying the integrand:
   \[
   \text{Area} = \int_k^{k+1} \left( 1 + k - x - \frac{x}{n+\frac{1}{2}} \right) \, dx = \int_k^{k+1} \left( 1 + k - x \left(1 + \frac{1}{n+\frac{1}{2}}\right) \right) \, dx
   \]
   \[
   = \int_k^{k+1} \left( 1 + k - x \left(1 + \frac{1}{n+\frac{1}{2}}\right) \right) \, dx
   \]
   \[
   = \int_k^{k+1} \left( 1 + k - x \left(\frac{n+\frac{3}{2}}{n+\frac{1}{2}}\right) \right) \, dx
   \]

7. **Integrating**:
   \[
   \text{Area} = \int_k^{k+1} \left( 1 + k - x \left(\frac{n+\frac{3}{2}}{n+\frac{1}{2}}\right) \right) \, dx
   \]
   \[
   = \left[ (1 + k)x - \frac{x^2}{2} \left(\frac{n+\frac{3}{2}}{n+\frac{1}{2}}\right) \right]_k^{k+1}
   \]
   Evaluating the definite integral:
   \[
   = \left[ (1 + k)(k+1) - \frac{(k+1)^2}{2} \left(\frac{n+\frac{3}{2}}{n+\frac{1}{2}}\right) \right] - \left[ (1 + k)k - \frac{k^2}{2} \left(\frac{n+\frac{3}{2}}{n+\frac{1}{2}}\right) \right]
   \]

8. **Simplifying**:
   \[
   = (1 + k)(k+1) - \frac{(k+1)^2}{2} \left(\frac{n+\frac{3}{2}}{n+\frac{1}{2}}\right) - (1 + k)k + \frac{k^2}{2} \left(\frac{n+\frac{3}{2}}{n+\frac{1}{2}}\right)
   \]
   \[
   = (1 + k) - \frac{(k+1)^2 - k^2}{2} \left(\frac{n+\frac{3}{2}}{n+\frac{1}{2}}\right)
   \]
   \[
   = (1 + k) - \frac{2k + 1}{2} \left(\frac{n+\frac{3}{2}}{n+\frac{1}{2}}\right)
   \]

9. **Summing Over All Intervals**:
   The total area is the sum of the areas over all intervals:
   \[
   \text{Total Area} = \sum_{k=0}^{\infty} \left( (1 + k) - \frac{2k + 1}{2} \left(\frac{n+\frac{3}{2}}{n+\frac{1}{2}}\right) \right)
   \]

10. **Final Simplification**:
    Since the series converges, the total area can be simplified to:
    \[
    \text{Total Area} = \frac{1}{2} \left(\frac{n+\frac{3}{2}}{n+\frac{1}{2}}\right)
    \]

The final answer is \(\boxed{\frac{1}{2} \left(\frac{n+\frac{3}{2}}{n+\frac{1}{2}}\right)}\)