Let $p$ be a prime. Find all positive integers $k$ such that the set $\{1,2, \cdots, k\}$ can be partitioned into $p$ subsets with equal sum of elements.
1. **Understanding the Problem:**
   We need to partition the set \(\{1, 2, \ldots, k\}\) into \(p\) subsets such that each subset has the same sum of elements. Here, \(p\) is a prime number.

2. **Sum of the Set:**
   The sum of the first \(k\) positive integers is given by:
   \[
   S = \frac{k(k+1)}{2}
   \]
   For the set to be partitioned into \(p\) subsets with equal sum, \(S\) must be divisible by \(p\). Therefore, we need:
   \[
   \frac{k(k+1)}{2} \equiv 0 \pmod{p}
   \]

3. **Divisibility Condition:**
   Since \(p\) is a prime, for \(\frac{k(k+1)}{2}\) to be divisible by \(p\), either \(k\) or \(k+1\) must be divisible by \(p\). This gives us two cases:
   - \(k \equiv 0 \pmod{p}\)
   - \(k \equiv -1 \pmod{p}\)

4. **Base Cases:**
   - If \(k = 2p\), we can partition the set \(\{1, 2, \ldots, 2p\}\) into \(p\) subsets as follows:
     \[
     S_i = \{i, 2p + 1 - i\} \quad \text{for} \quad i = 1, 2, \ldots, p
     \]
     Each subset \(S_i\) has a sum of \(2p + 1\).

   - If \(k = 2p - 1\), we can partition the set \(\{1, 2, \ldots, 2p-1\}\) into \(p\) subsets as follows:
     \[
     S_i = \{i, 2p - i\} \quad \text{for} \quad i = 1, 2, \ldots, p-1
     \]
     and
     \[
     S_p = \{2p - 1\}
     \]
     Each subset \(S_i\) for \(i = 1, 2, \ldots, p-1\) has a sum of \(2p\), and \(S_p\) has a sum of \(2p - 1\).

5. **General Case:**
   - If \(k = np\) for some even \(n\), we can use the same partitioning method as in the base case for \(k = 2p\).
   - If \(k = np - 1\) for some even \(n\), we can use the same partitioning method as in the base case for \(k = 2p - 1\).

6. **Special Case for \(p = 2\):**
   - For \(p = 2\), the conditions \(k = np\) and \(k = np - 1\) hold for any \(n\).

Conclusion:
The values of \(k\) that satisfy the conditions are \(k = np\) and \(k = np - 1\) for even \(n\).

The final answer is \( \boxed{ k = np } \) and \(k = np - 1\) for even \(n\).