To solve the problem \( 2n - 3 \) is divisible by \( n - 1 \), we can set up the equation as follows:

We need to determine when \( 2n - 3 \equiv 0 \mod (n - 1) \).

1. **Find the expression \( 2n - 3 \) in terms of \( n - 1 \):**

Let's express \( n \) in terms of \( k \):
\[
n = k + 1
\]
Substitute \( n \) into \( 2n - 3 \):
\[
2(k + 1) - 3 = 2k + 2 - 3 = 2k - 1
\]

2. **Now, we need to check if \( 2k - 1 \) is divisible by \( k \):**
\[
2k - 1 \equiv 0 \mod k
\]
This simplifies to checking:
\[
-1 \equiv 0 \mod k
\]

This means \( k \) must be a divisor of \( 1 \). The only positive divisor of \( 1 \) is \( 1 \), leading to:
\[
k = 1 \quad \Rightarrow \quad n = 2
\]

3. **Verification:**
Check if \( n = 2 \):
\[
2n - 3 = 2(2) - 3 = 4 - 3 = 1
\]
\[
n - 1 = 2 - 1 = 1
\]
\( 1 \) is indeed divisible by \( 1 \).

Thus, the only positive integer \( n \) that satisfies the condition is \( n = 2 \).