$ x$ and $ y$ are two distinct positive integers. What is the minimum positive integer value of $ (x \plus{} y^2)(x^2 \minus{} y)/(xy)$ ?

$\textbf{(A)}\ 3  \qquad\textbf{(B)}\ 8  \qquad\textbf{(C)}\ 14  \qquad\textbf{(D)}\ 15  \qquad\textbf{(E)}\ 17$
To find the minimum positive integer value of the expression \(\frac{(x + y^2)(x^2 - y)}{xy}\) where \(x\) and \(y\) are distinct positive integers, we will analyze the expression step-by-step.

1. **Rewrite the expression:**
   \[
   \frac{(x + y^2)(x^2 - y)}{xy}
   \]

2. **Expand the numerator:**
   \[
   (x + y^2)(x^2 - y) = x \cdot x^2 - x \cdot y + y^2 \cdot x^2 - y^2 \cdot y = x^3 - xy + x^2 y^2 - y^3
   \]

3. **Rewrite the expanded form:**
   \[
   \frac{x^3 - xy + x^2 y^2 - y^3}{xy}
   \]

4. **Separate the terms:**
   \[
   \frac{x^3}{xy} - \frac{xy}{xy} + \frac{x^2 y^2}{xy} - \frac{y^3}{xy} = \frac{x^2}{y} - 1 + xy - \frac{y^2}{x}
   \]

5. **Simplify the expression:**
   \[
   \frac{x^2}{y} - 1 + xy - \frac{y^2}{x}
   \]

6. **Consider specific values for \(x\) and \(y\):**
   - Let \(x = 2\) and \(y = 1\):
     \[
     \frac{(2 + 1^2)(2^2 - 1)}{2 \cdot 1} = \frac{(2 + 1)(4 - 1)}{2} = \frac{3 \cdot 3}{2} = \frac{9}{2} = 4.5 \quad \text{(not an integer)}
     \]
   - Let \(x = 3\) and \(y = 1\):
     \[
     \frac{(3 + 1^2)(3^2 - 1)}{3 \cdot 1} = \frac{(3 + 1)(9 - 1)}{3} = \frac{4 \cdot 8}{3} = \frac{32}{3} \quad \text{(not an integer)}
     \]
   - Let \(x = 4\) and \(y = 2\):
     \[
     \frac{(4 + 2^2)(4^2 - 2)}{4 \cdot 2} = \frac{(4 + 4)(16 - 2)}{8} = \frac{8 \cdot 14}{8} = 14 \quad \text{(integer)}
     \]

7. **Verify if 14 is the minimum:**
   - Let \(x = 5\) and \(y = 2\):
     \[
     \frac{(5 + 2^2)(5^2 - 2)}{5 \cdot 2} = \frac{(5 + 4)(25 - 2)}{10} = \frac{9 \cdot 23}{10} = \frac{207}{10} = 20.7 \quad \text{(not an integer)}
     \]
   - Let \(x = 3\) and \(y = 2\):
     \[
     \frac{(3 + 2^2)(3^2 - 2)}{3 \cdot 2} = \frac{(3 + 4)(9 - 2)}{6} = \frac{7 \cdot 7}{6} = \frac{49}{6} \quad \text{(not an integer)}
     \]

From the above calculations, the minimum positive integer value of the expression is indeed 14.

The final answer is \(\boxed{14}\).