Let $a$ be some positive number. Find the number of integer solutions $x$ of inequality $2 < xa < 3$ given that inequality $1 < xa < 2$ has exactly $3$ integer solutions. Consider all possible cases.

[i](4 points)[/i]

1. **Understanding the given conditions:**
   - We are given that the inequality \(1 < xa < 2\) has exactly 3 integer solutions.
   - We need to find the number of integer solutions for the inequality \(2 < xa < 3\).

2. **Analyzing the first inequality \(1 < xa < 2\):**
   - Let \(x\) be an integer. For \(1 < xa < 2\) to have 3 integer solutions, \(a\) must be such that there are exactly 3 integers \(x\) satisfying this inequality.
   - Let \(x_1, x_2, x_3\) be the three integer solutions. Then:
     \[
     1 < x_1 a < 2, \quad 1 < x_2 a < 2, \quad 1 < x_3 a < 2
     \]
   - Since \(x_1, x_2, x_3\) are consecutive integers (because \(a\) is a positive number and the range \(1 < xa < 2\) is small), we can denote them as \(x, x+1, x+2\).

3. **Finding the range of \(a\):**
   - For \(x\) to be an integer solution, we have:
     \[
     1 < xa < 2 \implies \frac{1}{x} < a < \frac{2}{x}
     \]
   - For \(x+1\) to be an integer solution, we have:
     \[
     1 < (x+1)a < 2 \implies \frac{1}{x+1} < a < \frac{2}{x+1}
     \]
   - For \(x+2\) to be an integer solution, we have:
     \[
     1 < (x+2)a < 2 \implies \frac{1}{x+2} < a < \frac{2}{x+2}
     \]
   - Since \(a\) must satisfy all three inequalities simultaneously, we take the intersection of these intervals:
     \[
     \max\left(\frac{1}{x}, \frac{1}{x+1}, \frac{1}{x+2}\right) < a < \min\left(\frac{2}{x}, \frac{2}{x+1}, \frac{2}{x+2}\right)
     \]
   - For \(a\) to be valid, it must lie in the intersection of these ranges. Given that there are exactly 3 integer solutions, we can infer that:
     \[
     \frac{1}{x+2} < a < \frac{2}{x}
     \]

4. **Determining the range of \(a\):**
   - Since \(a\) must lie between \(\frac{1}{x+2}\) and \(\frac{2}{x}\), and given that there are exactly 3 integer solutions, we can deduce that \(x\) must be such that:
     \[
     \frac{1}{x+2} < a < \frac{2}{x}
     \]
   - To have 3 integer solutions, \(a\) must be in the range where \(x, x+1, x+2\) are valid. This implies:
     \[
     \frac{1}{4} < a < \frac{1}{3}
     \]

5. **Analyzing the second inequality \(2 < xa < 3\):**
   - We now need to find the number of integer solutions for \(2 < xa < 3\).
   - Using the same logic as above, we have:
     \[
     2 < xa < 3 \implies \frac{2}{x} < a < \frac{3}{x}
     \]
   - For \(x+1\) to be an integer solution, we have:
     \[
     2 < (x+1)a < 3 \implies \frac{2}{x+1} < a < \frac{3}{x+1}
     \]
   - For \(x+2\) to be an integer solution, we have:
     \[
     2 < (x+2)a < 3 \implies \frac{2}{x+2} < a < \frac{3}{x+2}
     \]
   - Since \(a\) must satisfy all three inequalities simultaneously, we take the intersection of these intervals:
     \[
     \max\left(\frac{2}{x}, \frac{2}{x+1}, \frac{2}{x+2}\right) < a < \min\left(\frac{3}{x}, \frac{3}{x+1}, \frac{3}{x+2}\right)
     \]

6. **Determining the number of integer solutions:**
   - Given that \(a\) lies between \(\frac{1}{4}\) and \(\frac{1}{3}\), we need to check if the same \(a\) will provide 3 integer solutions for the inequality \(2 < xa < 3\).
   - Since the range of \(a\) is the same, the number of integer solutions for \(2 < xa < 3\) will also be 3.

Conclusion:
\[
\boxed{3}
\]