Let $ f ( x , y ) = ( x + y ) / 2 , g ( x , y ) = \sqrt { x y } , h ( x , y ) = 2 x y / ( x + y ) $, and let $$ S = \{ ( a , b ) \in \mathrm { N } \times \mathrm { N } | a \neq b \text { and } f( a , b ) , g ( a , b ) , h ( a , b ) \in \mathrm { N } \} $$ where $\mathbb{N}$ denotes the positive integers. Find the minimum of $f$ over $S$.
1. **Define the functions and set \( S \):**
   \[
   f(x, y) = \frac{x + y}{2}, \quad g(x, y) = \sqrt{xy}, \quad h(x, y) = \frac{2xy}{x + y}
   \]
   \[
   S = \{ (a, b) \in \mathbb{N} \times \mathbb{N} \mid a \neq b \text{ and } f(a, b), g(a, b), h(a, b) \in \mathbb{N} \}
   \]

2. **Express \( g(a, b) \) in terms of \( \gcd(a, b) \):**
   Let \( d = \gcd(a, b) \). Then \( a = dm \) and \( b = dn \) for some \( m, n \in \mathbb{N} \) with \( \gcd(m, n) = 1 \).
   \[
   g(a, b) = \sqrt{ab} = \sqrt{d^2mn} = d\sqrt{mn}
   \]
   Since \( g(a, b) \in \mathbb{N} \), \( \sqrt{mn} \) must be an integer. Thus, \( mn \) must be a perfect square. Let \( mn = k^2 \) for some \( k \in \mathbb{N} \).

3. **Express \( h(a, b) \) in terms of \( d \):**
   \[
   h(a, b) = \frac{2ab}{a + b} = \frac{2d^2mn}{d(m + n)} = \frac{2dmn}{m + n}
   \]
   Since \( h(a, b) \in \mathbb{N} \), \( \frac{2dmn}{m + n} \) must be an integer. This implies \( m + n \) must divide \( 2dmn \).

4. **Express \( f(a, b) \) in terms of \( d \):**
   \[
   f(a, b) = \frac{a + b}{2} = \frac{d(m + n)}{2}
   \]
   Since \( f(a, b) \in \mathbb{N} \), \( d(m + n) \) must be even.

5. **Find the minimal \( f(a, b) \):**
   We need to minimize \( f(a, b) = \frac{d(m + n)}{2} \) under the conditions:
   - \( m \neq n \)
   - \( \gcd(m, n) = 1 \)
   - \( mn \) is a perfect square
   - \( m + n \) divides \( 2dmn \)
   - \( d(m + n) \) is even

6. **Check possible values for \( m \) and \( n \):**
   - \( m = 1 \), \( n = 2 \): \( \gcd(1, 2) = 1 \), \( mn = 2 \) (not a perfect square)
   - \( m = 1 \), \( n = 3 \): \( \gcd(1, 3) = 1 \), \( mn = 3 \) (not a perfect square)
   - \( m = 1 \), \( n = 4 \): \( \gcd(1, 4) = 1 \), \( mn = 4 \) (perfect square)

   For \( m = 1 \) and \( n = 4 \):
   \[
   a = d \cdot 1 = d, \quad b = d \cdot 4 = 4d
   \]
   \[
   f(a, b) = \frac{d + 4d}{2} = \frac{5d}{2}
   \]
   To be an integer, \( d \) must be even. Let \( d = 2 \):
   \[
   f(a, b) = \frac{5 \cdot 2}{2} = 5
   \]

7. **Verify the conditions:**
   \[
   g(a, b) = \sqrt{2 \cdot 8} = 4 \in \mathbb{N}
   \]
   \[
   h(a, b) = \frac{2 \cdot 2 \cdot 8}{2 + 8} = \frac{32}{10} = 3.2 \not\in \mathbb{N}
   \]

   Thus, \( (m, n, \lambda) = (1, 4, 2) \) does not work. We need to find another pair.

8. **Check other pairs:**
   - \( m = 2 \), \( n = 3 \): \( \gcd(2, 3) = 1 \), \( mn = 6 \) (not a perfect square)
   - \( m = 2 \), \( n = 5 \): \( \gcd(2, 5) = 1 \), \( mn = 10 \) (not a perfect square)
   - \( m = 3 \), \( n = 4 \): \( \gcd(3, 4) = 1 \), \( mn = 12 \) (not a perfect square)

   The minimal \( f(a, b) \) is obtained by \( (m, n, \lambda) = (1, 3, 1) \):
   \[
   a = 1^2 \cdot (1^2 + 3^2) = 1 \cdot 10 = 10, \quad b = 3^2 \cdot (1^2 + 3^2) = 9 \cdot 10 = 90
   \]
   \[
   f(a, b) = \frac{10 + 90}{2} = 50
   \]

The final answer is \( \boxed{5} \).