Find the number of sets $A$ that satisfy the three conditions:
$\star$ $A$ is a set of two positive integers
$\star$ each of the numbers in $A$ is at least $22$ percent the size of the other number
$\star$ $A$ contains the number $30.$
1. We start by noting that the set \( A \) must contain the number 30 and another positive integer \( x \). Therefore, \( A = \{30, x\} \).

2. The first condition states that each number in \( A \) must be at least 22% the size of the other number. This gives us two inequalities:
   - \( x \geq 0.22 \times 30 \)
   - \( 30 \geq 0.22 \times x \)

3. Solving the first inequality:
   \[
   x \geq 0.22 \times 30 \implies x \geq 6.6
   \]
   Since \( x \) must be an integer, we round up to the nearest whole number:
   \[
   x \geq 7
   \]

4. Solving the second inequality:
   \[
   30 \geq 0.22 \times x \implies \frac{30}{0.22} \geq x \implies 136.\overline{36} \geq x
   \]
   Since \( x \) must be an integer, we round down to the nearest whole number:
   \[
   x \leq 136
   \]

5. Combining these results, we have:
   \[
   7 \leq x \leq 136
   \]

6. The integers \( x \) can take values from 7 to 136 inclusive. To find the number of possible values for \( x \), we calculate:
   \[
   136 - 7 + 1 = 130
   \]

7. However, we must exclude the case where \( x = 30 \) because the set \( \{30, 30\} \) is not valid (it is not a set of two distinct integers).

8. Therefore, we subtract 1 from the total count:
   \[
   130 - 1 = 129
   \]

The final answer is \(\boxed{129}\).