Find all values of the positive integer $k$ that has the property:
There are no positive integers $a,b$ such that the expression $A(k,a,b)=\frac{a+b}{a^2+k^2b^2-k^2ab}$ is a composite positive number.
1. We start by analyzing the given expression \( A(k, a, b) = \frac{a + b}{a^2 + k^2 b^2 - k^2 ab} \).

2. We need to find all positive integer values of \( k \) such that there are no positive integers \( a \) and \( b \) for which \( A(k, a, b) \) is a composite positive number.

3. First, consider the case when \( k = 1 \):
   \[
   A(1, a, b) = \frac{a + b}{a^2 + b^2 - ab}
   \]

4. We will show that \( A(1, a, b) \leq 2 \):
   - For \( a = b \):
     \[
     A(1, a, a) = \frac{2a}{a^2} = \frac{2}{a}
     \]
     Since \( a \) is a positive integer, \( \frac{2}{a} \leq 2 \).

   - Without loss of generality, assume \( a < b \):
     \[
     A(1, a, b) = \frac{a + b}{a^2 + b^2 - ab}
     \]
     We need to show that this expression is less than or equal to 1:
     \[
     A(1, a, b) = \frac{a + b}{a^2 + b(b - a)}
     \]
     Since \( b > a \), we have:
     \[
     a^2 + b(b - a) = a^2 + b^2 - ab \geq a^2 + b - a
     \]
     Therefore:
     \[
     A(1, a, b) \leq \frac{a + b}{a + b} = 1
     \]

5. Hence, for \( k = 1 \), \( A(1, a, b) \) is always less than or equal to 2, and thus cannot be a composite number greater than 1.

6. Now consider \( k \geq 2 \):
   - For \( a = k^2 - 1 \) and \( b = 1 \):
     \[
     A(k, k^2 - 1, 1) = \frac{(k^2 - 1) + 1}{(k^2 - 1)^2 + k^2 - k^2(k^2 - 1)}
     \]
     Simplifying the denominator:
     \[
     (k^2 - 1)^2 + k^2 - k^2(k^2 - 1) = k^4 - 2k^2 + 1 + k^2 - k^4 + k^2 = 2k^2 - 1
     \]
     Therefore:
     \[
     A(k, k^2 - 1, 1) = \frac{k^2}{k^2} = k^2
     \]
     Since \( k \geq 2 \), \( k^2 \) is a composite number.

7. Therefore, for any \( k \geq 2 \), \( A(k, a, b) \) can be a composite number.

8. Hence, the only value of \( k \) that satisfies the given property is \( k = 1 \).

The final answer is \( \boxed{ k = 1 } \).