Find all positive integers $n$ for which there exists a set of exactly $n$ distinct positive integers, none of which exceed $n^2$, whose reciprocals add up to $1$.
1. **Define the set and the problem constraints:**
   Let \( S_n = \{a_1, a_2, \dots, a_n\} \) be a set of \( n \) distinct positive integers where \( a_1 < a_2 < \dots < a_n \le n^2 \) and \( \sum_{i=1}^n \frac{1}{a_i} = 1 \).

2. **Check for \( n = 1 \):**
   For \( n = 1 \), let \( S_1 = \{1\} \). Clearly, \( \frac{1}{1} = 1 \), so \( S_1 \) satisfies the condition.

3. **Check for \( n = 2 \):**
   For \( n = 2 \), we need \( \frac{1}{a_1} + \frac{1}{a_2} = 1 \). This implies:
   \[
   \frac{1}{a_1} + \frac{1}{a_2} = 1 \implies \frac{a_1 + a_2}{a_1 a_2} = 1 \implies a_1 a_2 = a_1 + a_2
   \]
   Rearranging, we get:
   \[
   a_1 a_2 - a_1 - a_2 = 0 \implies (a_1 - 1)(a_2 - 1) = 1
   \]
   The only integer solution to this equation is \( (a_1, a_2) = (2, 2) \), but \( a_1 \) and \( a_2 \) must be distinct, leading to a contradiction. Therefore, no such set \( S_2 \) exists.

4. **Check for \( n \ge 3 \):**
   We will show that \( S_n \) exists for \( n \ge 3 \).

   - **Case 1: \( n \) cannot be expressed as \( m^2 + m \) for some integer \( m \ge 2 \):**
     Let \( S_n = \{k(k+1) \mid 1 \le k \le n-1\} \cup \{n\} \). Then:
     \[
     \sum_{k=1}^{n-1} \frac{1}{k(k+1)} + \frac{1}{n} = \sum_{k=1}^{n-1} \left( \frac{1}{k} - \frac{1}{k+1} \right) + \frac{1}{n}
     \]
     This is a telescoping series:
     \[
     \left( 1 - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \cdots + \left( \frac{1}{n-1} - \frac{1}{n} \right) + \frac{1}{n} = 1
     \]
     Thus, \( S_n \) satisfies the condition.

   - **Case 2: \( n = m^2 + m \) for some integer \( m \ge 2 \):**
     Let \( S_n = \{k(k+1) \mid 2 \le k \le n-3\} \cup \{3, 8, 24, n-2\} \). Then:
     \[
     \sum_{k=2}^{n-3} \frac{1}{k(k+1)} + \frac{1}{3} + \frac{1}{8} + \frac{1}{24} + \frac{1}{n-2}
     \]
     The sum of the first part is:
     \[
     \sum_{k=2}^{n-3} \left( \frac{1}{k} - \frac{1}{k+1} \right) = \frac{1}{2} - \frac{1}{n-2}
     \]
     Adding the remaining terms:
     \[
     \frac{1}{2} - \frac{1}{n-2} + \frac{1}{3} + \frac{1}{8} + \frac{1}{24} + \frac{1}{n-2} = \frac{1}{2} + \frac{1}{2} = 1
     \]
     Thus, \( S_n \) satisfies the condition.

5. **Conclusion:**
   Since \( S_n \) exists for all \( n \ge 3 \) and \( n = 1 \), but not for \( n = 2 \), the answer is all positive integers \( n \neq 2 \).

The final answer is \(\boxed{\text{all positive integers} \: n \ne 2}\)