More generally, let $a,b,c,d,p,$ and $q$ be positive integers such that $bc-ad=1$ and \frac ab 0,$ or \[bp-aq\geq1. \hspace{15mm} (1)
From $\frac pq 0,$ or $$cq-dp\geq1. \hspace{15mm} (2)$$
Since $bc-ad=1,$ note that:

Multiplying $(1)$ by $d,$ multiplying $(2)$ by $b,$ and adding the results, we get $q\geq b+d.$
Multiplying $(1)$ by $c,$ multiplying $(2)$ by $a,$ and adding the results, we get $p\geq a+c.$

To minimize $q,$ we set $q=b+d,$ from which $p=a+c.$ Together, we can prove that \[\frac{a+c-1}{b+d}\leq\frac abk_1, \\ \frac{a+c}{b+d}&=\frac{bk_1+dk_2}{b+d}&&=k_2+\frac{bk_1-bk_2}{b+d}&&=k_2+\frac{b(k_1-k_2)}{b+d}&&<k_2. \end{alignat*}
Moreover, this part of $(\bigstar)$ is independent of the precondition $bc-ad=1.$

~MRENTHUSIASM

To solve the problem, we need to find the smallest positive integer $q$ such that there exists a positive integer $p$ satisfying the inequality:
$$\frac{5}{9} < \frac{p}{q} < \frac{4}{7}$$

1. **Express the inequalities in terms of $p$ and $q$:**
$$\frac{5}{9} < \frac{p}{q} < \frac{4}{7}$$
This can be rewritten as:
$$5q < 9p \quad \text{and} \quad 7p < 4q$$

2. **Manipulate the inequalities to find bounds for $q$:**
- From $5q < 9p$:
$$5q < 9p \implies p > \frac{5q}{9}$$
- From $7p < 4q$:
$$7p < 4q \implies p < \frac{4q}{7}$$

3. Combine the inequalities:
$$\frac{5q}{9} < p < \frac{4q}{7}$$

4. **Find the smallest $q$ such that there exists an integer $p$ satisfying the combined inequality:**
- We need to find the smallest $q$ such that:
$$\left\lceil \frac{5q}{9} \right\rceil < \left\lfloor \frac{4q}{7} \right\rfloor$$

5. **Test values of $q$ starting from the smallest possible integer:**
- For $q = 1$:
$$\left\lceil \frac{5 \cdot 1}{9} \right\rceil = \left\lceil \frac{5}{9} \right\rceil = 1, \quad \left\lfloor \frac{4 \cdot 1}{7} \right\rfloor = \left\lfloor \frac{4}{7} \right\rfloor = 0 \quad \text{(not valid)}$$
- For $q = 2$:
$$\left\lceil \frac{5 \cdot 2}{9} \right\rceil = \left\lceil \frac{10}{9} \right\rceil = 2, \quad \left\lfloor \frac{4 \cdot 2}{7} \right\rfloor = \left\lfloor \frac{8}{7} \right\rfloor = 1 \quad \text{(not valid)}$$
- Continue this process until $q = 16$:
$$\left\lceil \frac{5 \cdot 16}{9} \right\rceil = \left\lceil \frac{80}{9} \right\rceil = 9, \quad \left\lfloor \frac{4 \cdot 16}{7} \right\rfloor = \left\lfloor \frac{64}{7} \right\rfloor = 9 \quad \text{(valid)}$$

6. **Verify the values of $p$ and $q$:**
- For $q = 16$ and $p = 9$:
$$\frac{5}{9} < \frac{9}{16} < \frac{4}{7}$$
- Check:
$$\frac{5}{9} \approx 0.5556, \quad \frac{9}{16} = 0.5625, \quad \frac{4}{7} \approx 0.5714$$
- The inequality holds true.

7. **Calculate $q - p$:**
$$q - p = 16 - 9 = 7$$

The final answer is $\boxed{7}$