Let there be positive integers $a, c$. Positive integer $b$ is a divisor of $ac-1$. For a positive rational number $r$ which is less than $1$, define the set $A(r)$ as follows. 
$$A(r) = \{m(r-ac)+nab| m, n \in \mathbb{Z} \}$$
Find all rational numbers $r$ which makes the minimum positive rational number in $A(r)$ greater than or equal to $\frac{ab}{a+b}$.
1. Let \( r = \frac{p}{q} \) for coprime positive integers \( p < q \).

2. Consider the expression \( m(r - ac) + nab \):
   \[
   m\left(\frac{p}{q} - ac\right) + nab = \frac{m(p - qac) + n(qab)}{q}
   \]

3. Note that the set \( \{ mx + ny \mid m, n \in \mathbb{Z} \} = \{ k \gcd(x, y) \mid k \in \mathbb{Z} \} \). Therefore:
   \[
   A(r) = \left\{ \frac{k \cdot \gcd(p - qac, qab)}{q} \mid k \in \mathbb{Z} \right\}
   \]

4. To find the minimum positive element of \( A(r) \), we need to compute \( \gcd(p - qac, qab) \). Since \( \gcd(p - qac, q) = \gcd(p, q) = 1 \), we have:
   \[
   \gcd(p - qac, qab) = \gcd(p - qac, ab)
   \]

5. Given \( b \mid ac - 1 \), we have \( \gcd(b, ac) \mid \gcd(ac - 1, ac) = 1 \). Thus, \( \gcd(b, ac) = 1 \), which implies \( \gcd(b, a) = 1 \).

6. Therefore:
   \[
   \gcd(p - qac, ab) = \gcd(p - qac, a) \cdot \gcd(p - qac, b) = \gcd(p, a) \cdot \gcd(q - p, b)
   \]

7. The minimum positive element of \( A(r) \) is:
   \[
   \frac{\gcd(p - qac, qab)}{q} = \frac{\gcd(p, a) \cdot \gcd(q - p, b)}{q}
   \]

8. Let \( p = gu \), \( a = gv \), \( q - p = g'u' \), and \( b = g'v' \), where \( g = \gcd(p, a) \), \( g' = \gcd(q - p, b) \), and \( g, g', u, u', v, v' \geq 1 \). Then:
   \[
   \frac{\gcd(p, a) \cdot \gcd(q - p, b)}{q} = \frac{gg'}{gu + g'u'}
   \]

9. We need this to be greater than or equal to \( \frac{ab}{a + b} \):
   \[
   \frac{gg'}{gu + g'u'} \geq \frac{gg'vv'}{gv + g'v'}
   \]

10. Simplifying, we get:
    \[
    gu + g'u' \leq \frac{g}{v'} + \frac{g'}{v}
    \]

11. Since \( g + g' \leq gu + g'u' \leq \frac{g}{v'} + \frac{g'}{v} \leq g + g' \), it follows that \( u = u' = v = v' = 1 \). Therefore, \( p = g = a \) and \( q - p = g' = b \).

12. Hence, the only \( r \) that satisfies the condition is:
    \[
    r = \frac{a}{a + b}
    \]

The final answer is \( \boxed{ \frac{a}{a + b} } \)