Let $n$ be a fixed positive integer. There are $n \geq 1$ lamps in a row, some of them are on and some are off. In a single move, we choose a positive integer $i$ ($1 \leq i \leq n$) and switch the state of the first $i$ lamps from the left. Determine the smallest number $k$ with the property that we can make all of the lamps be switched on using at most $k$ moves, no matter what the initial configuration was.

[i]Proposed by Viktor Simjanoski and Nikola Velov[/i]
1. **Understanding the Problem:**
   - We have \( n \) lamps in a row, each of which can be either on or off.
   - In a single move, we can choose a positive integer \( i \) (where \( 1 \leq i \leq n \)) and switch the state (on to off or off to on) of the first \( i \) lamps from the left.
   - We need to determine the smallest number \( k \) such that we can make all the lamps be switched on using at most \( k \) moves, regardless of the initial configuration.

2. **Analyzing the Moves:**
   - Let's denote the state of the lamps as a binary string of length \( n \), where 1 represents a lamp that is on and 0 represents a lamp that is off.
   - The goal is to transform any initial configuration into a string of all 1s (i.e., all lamps on).

3. **Strategy to Turn All Lamps On:**
   - Consider the initial configuration of the lamps. We can use a sequence of moves to turn all lamps on.
   - If the \( i \)-th lamp is off (0), we can choose \( i \) and switch the state of the first \( i \) lamps. This will turn the \( i \)-th lamp on (1) and potentially change the state of the previous \( i-1 \) lamps.

4. **Inductive Approach:**
   - We can use induction to show that \( k = n \) moves are sufficient.
   - **Base Case:** For \( n = 1 \), if the lamp is off, we choose \( i = 1 \) and switch it on. This requires 1 move.
   - **Inductive Step:** Assume that for \( n = k \), we can turn all lamps on using at most \( k \) moves. For \( n = k+1 \), consider the initial configuration of \( k+1 \) lamps. If the last lamp (the \((k+1)\)-th lamp) is off, we can choose \( i = k+1 \) and switch the state of the first \( k+1 \) lamps. This will turn the \((k+1)\)-th lamp on and change the state of the first \( k \) lamps. By the inductive hypothesis, we can turn the first \( k \) lamps on using at most \( k \) moves. Thus, we need at most \( k+1 \) moves for \( n = k+1 \).

5. **Conclusion:**
   - By induction, we have shown that \( k = n \) moves are sufficient to turn all lamps on for any initial configuration.
   - Since each move can only switch the state of the first \( i \) lamps, and we need to ensure that each lamp is turned on at least once, \( k = n \) is also necessary.

The final answer is \( \boxed{ k = n } \).