There are $n$ lamps in a row. Some of which are on. Every minute all the lamps already on go off. Those which were off and were adjacent to exactly one lamp which was on will go on. For which $n$ one can find an initial configuration of lamps which were on, such that at least one lamp will be on at any time?
To determine for which values of \( n \) one can find an initial configuration of lamps such that at least one lamp will be on at any time, we need to analyze the given configurations and verify their correctness.

1. **Case \( n = 1 \):**
   - If there is only one lamp, it will turn off in the first minute and there will be no adjacent lamps to turn it back on. Therefore, \( n = 1 \) is not a solution.

2. **Case \( n = 2 \):**
   - Initial configuration: \( 01 \)
   - After 1 minute: \( 10 \)
   - After 2 minutes: \( 01 \)
   - The lamps alternate between \( 01 \) and \( 10 \), ensuring that at least one lamp is always on. Therefore, \( n = 2 \) is a solution.

3. **Case \( n = 3 \):**
   - Any initial configuration will result in all lamps turning off after a few minutes. For example:
     - Initial configuration: \( 010 \)
     - After 1 minute: \( 101 \)
     - After 2 minutes: \( 010 \)
     - After 3 minutes: \( 000 \)
   - Therefore, \( n = 3 \) is not a solution.

4. **Case \( n = 4k \):**
   - Initial configuration: \( 0110 \) repeated \( k \) times.
   - For \( k = 1 \), the configuration is \( 0110 \):
     - After 1 minute: \( 1001 \)
     - After 2 minutes: \( 0110 \)
     - The pattern repeats, ensuring that at least one lamp is always on.
   - For larger \( k \), the same pattern holds. Therefore, \( n = 4k \) is a solution.

5. **Case \( n = 4k + 1 \):**
   - Initial configuration: \( 0110 \) repeated \( k-1 \) times followed by \( 01010 \).
   - For \( k = 1 \), the configuration is \( 01010 \):
     - After 1 minute: \( 10101 \)
     - After 2 minutes: \( 01010 \)
     - The pattern repeats, ensuring that at least one lamp is always on.
   - For larger \( k \), the same pattern holds. Therefore, \( n = 4k + 1 \) is a solution.

6. **Case \( n = 4k + 2 \):**
   - Initial configuration: \( 0110 \) repeated \( k \) times followed by \( 01 \).
   - For \( k = 1 \), the configuration is \( 011001 \):
     - After 1 minute: \( 100110 \)
     - After 2 minutes: \( 011001 \)
     - The pattern repeats, ensuring that at least one lamp is always on.
   - For larger \( k \), the same pattern holds. Therefore, \( n = 4k + 2 \) is a solution.

7. **Case \( n = 4k + 3 \):**
   - Initial configuration: \( 0110 \) repeated \( k \) times followed by \( 001 \).
   - For \( k = 1 \), the configuration is \( 0110001 \):
     - After 1 minute: \( 1001000 \)
     - After 2 minutes: \( 0110001 \)
     - The pattern repeats, ensuring that at least one lamp is always on.
   - For larger \( k \), the same pattern holds. Therefore, \( n = 4k + 3 \) is a solution.

The final answer is \( \boxed{ n \neq 1, 3 } \).