Let $n$ be a positive integer. There are $n$ lamps, each with a switch that changes the lamp from on to off, or from off to on, each time it is pressed. The lamps are initially all off.
You are going to press the switches in a series of rounds. In the first round, you are going to press exactly $1$ switch; in the second round, you are going to press exactly $2$ switches; and so on, so that in the $k$th round you are going to press exactly $k$ switches. In each round you will press each switch at most once. Your goal is to finish a round with all of the lamps switched on.
Determine for which $n$ you can achieve this goal.
To determine for which positive integers \( n \) it is possible to achieve the goal of having all lamps switched on after a series of rounds, we need to analyze the switching process in detail.

1. **Base Cases:**
   - For \( n = 1 \):
     - In the first round, we press the only switch, turning the lamp on. Thus, it is possible to achieve the goal.
   - For \( n = 2 \):
     - In the first round, we press one switch, turning one lamp on.
     - In the second round, we press both switches, changing the state of both lamps. This will result in one lamp being on and the other being off, or both being off again. Thus, it is not possible to achieve the goal for \( n = 2 \).
   - For \( n = 3 \):
     - In the first round, we press one switch, turning one lamp on.
     - In the second round, we press two switches, changing the state of two lamps.
     - In the third round, we press all three switches, changing the state of all lamps. This will result in all lamps being on. Thus, it is possible to achieve the goal for \( n = 3 \).

2. **General Case Analysis:**
   - We index the lamps \( 1, 2, \dots, n \).
   - Consider the following "standard process":
     - In round \( k \), we switch \( k \) lamps. If \( k \) is odd, we switch lamps \( 2, \dots, k+1 \). If \( k \) is even, we switch lamps \( 1, \dots, k \).
     - The effect of round \( 2k-1 \) and round \( 2k \) combined is equivalent to switching just lamp \( 1 \).

3. **Even \( n \):**
   - If \( n = 2m \) is even, we choose an even integer \( k \) such that \( m \leq 2k \leq n \). This is possible whenever \( n \geq 4 \).
   - We switch the lamps according to the standard process, except that in round \( m \), we switch only the lamps that are not switched in the standard process.
     - If \( m \) is odd, we switch lamps \( 1, m+2, m+3, \dots, 2m \).
     - If \( m \) is even, we switch lamps \( m+1, m+2, \dots, 2m \).
   - After \( 2k \) rounds, the standard process will end with all lamps being off, but our modified process will be equivalent to all lamps being switched once more, thus all on.

4. **Odd \( n \):**
   - If \( n = 4m + 1 \):
     - We switch lamps according to the standard process, except that in round \( n \), we switch all lamps \( 1, \dots, n \).
     - After round \( 4m \), all lamps will be off. Therefore, after round \( n \), all lamps are on.
   - If \( n = 4m + 3 \):
     - We choose an odd integer \( k \) such that \( 2m + 2 \leq 2k \leq n \). This is possible whenever \( n \geq 7 \).
     - We switch the lamps according to the standard process, except that in round \( 2m + 1 \), we switch lamps \( 2m + 3, \dots, 4m + 3 \).
     - After \( 2k \) rounds, the standard process will end with only lamp \( 1 \) on. Our modified process will be equivalent to lamps \( 2, \dots, 4m + 3 \) being switched once more, thus all lamps are on.

Conclusion:
- The goal can be achieved for all positive integers \( n \) except \( n = 2 \).

The final answer is all positive integers except \( \boxed{ n = 2 } \).