Let $ a > b > 1$ be relatively prime positive integers. Define the weight of an integer $ c$, denoted by $ w(c)$ to be the minimal possible value of $ |x| \plus{} |y|$ taken over all pairs of integers $ x$ and $ y$ such that \[ax \plus{} by \equal{} c.\] An integer $ c$ is called a [i]local champion [/i]if $ w(c) \geq w(c \pm a)$ and $ w(c) \geq w(c \pm b)$. 

Find all local champions and determine their number.

[i]Proposed by Zoran Sunic, USA[/i]
1. **Define the problem and initial conditions:**
   Let \( a > b > 1 \) be relatively prime positive integers. Define the weight of an integer \( c \), denoted by \( w(c) \), to be the minimal possible value of \( |x| + |y| \) taken over all pairs of integers \( x \) and \( y \) such that \( ax + by = c \). An integer \( c \) is called a *local champion* if \( w(c) \geq w(c \pm a) \) and \( w(c) \geq w(c \pm b) \).

2. **Express \( c \) in terms of \( a \) and \( b \):**
   Suppose \( c = ax + by \) for some integers \( x \) and \( y \). Since \( a \) and \( b \) are relatively prime, by the Extended Euclidean Algorithm, there exist integers \( x \) and \( y \) such that \( ax + by = c \).

3. **Transform the equation:**
   Notice that if \( c = ax + by \), then for any integer \( k \), we have:
   \[
   c = a(x - bk) + b(y + ak)
   \]
   This transformation allows us to explore different pairs \((x, y)\) that satisfy the equation.

4. **Consider the weight function \( w(c) \):**
   The weight \( w(c) \) is defined as the minimal value of \( |x| + |y| \) for all pairs \((x, y)\) such that \( ax + by = c \).

5. **Analyze the local champion condition:**
   For \( c \) to be a local champion, it must satisfy:
   \[
   w(c) \geq w(c + a), \quad w(c) \geq w(c - a), \quad w(c) \geq w(c + b), \quad w(c) \geq w(c - b)
   \]

6. **Consider the minimal weight configuration:**
   Without loss of generality, assume \( c = ax - by \) where \( x \) and \( y \) have the same sign (both positive). This is because if \( x \) and \( y \) have opposite signs, we can always find a pair with smaller absolute values.

7. **Explore the transformation for minimal weight:**
   Consider the transformation \( c = a(x - bk) + b(y + ak) \). We need to find the minimal \( k \) such that the weight \( |x - bk| + |y + ak| \) is minimized.

8. **Determine the minimal \( k \):**
   By examining the cases where \( y < ak \) and other similar conditions, we find that the minimal \( k \) should be 1. This is because any larger \( k \) would increase the weight.

9. **Identify the local champions:**
   The only local champions are those for which \( x + y = \left\lfloor \frac{a + b}{2} \right\rfloor \). This is because this configuration minimizes the weight while satisfying the local champion conditions.

10. **Verify the solution:**
    By checking the conditions for local champions, we confirm that the integers \( c \) for which \( x + y = \left\lfloor \frac{a + b}{2} \right\rfloor \) indeed satisfy \( w(c) \geq w(c \pm a) \) and \( w(c) \geq w(c \pm b) \).

The final answer is \( \boxed{ c } \) such that \( x + y = \left\lfloor \frac{a + b}{2} \right\rfloor \).