On the control board of a nuclear station, there are $n$ electric switches ($n > 0$), all in one row. Each switch has two possible positions: up and down. The switches are connected to each other in such a way that, whenever a switch moves down from its upper position, its right neighbour (if it exists) automatically changes position. At the beginning, all switches are down. The operator of the board first changes the position of the leftmost switch once, then the position of the second leftmost switch twice etc., until eventually he changes the position of the rightmost switch n times. How many switches are up after all these operations?
1. **Initial Setup**: 
   - We have \( n \) switches, all initially in the down position.
   - Each switch has two possible positions: up and down.
   - When a switch moves down from its upper position, its right neighbor (if it exists) automatically changes position.

2. **Operations**:
   - The operator changes the position of the leftmost switch once, the second leftmost switch twice, and so on, until the rightmost switch is changed \( n \) times.

3. **Analysis of Switch Changes**:
   - Let's denote the switches as \( S_1, S_2, \ldots, S_n \).
   - The number of times each switch \( S_i \) is toggled is as follows:
     - \( S_1 \) is toggled 1 time.
     - \( S_2 \) is toggled 2 times.
     - \( S_3 \) is toggled 3 times.
     - ...
     - \( S_n \) is toggled \( n \) times.

4. **Effect of Toggling**:
   - A switch toggled an odd number of times will end up in the up position.
   - A switch toggled an even number of times will end up in the down position.

5. **Determining the Final Position**:
   - We need to determine how many switches are toggled an odd number of times.
   - The number of times each switch \( S_i \) is toggled is equal to the number of divisors of \( i \) (since each divisor corresponds to a toggle operation).

6. **Odd Number of Divisors**:
   - A number has an odd number of divisors if and only if it is a perfect square. This is because divisors generally come in pairs, except for perfect squares which have a middle divisor that is repeated (e.g., \( 4 = 2 \times 2 \)).

7. **Counting Perfect Squares**:
   - The number of perfect squares between 1 and \( n \) is the number of integers \( k \) such that \( k^2 \leq n \).
   - This is equivalent to finding the largest integer \( k \) such that \( k \leq \sqrt{n} \).

8. **Conclusion**:
   - The number of switches that are up after all operations is equal to the number of perfect squares less than or equal to \( n \).
   - This number is \( \lfloor \sqrt{n} \rfloor \).

The final answer is \( \boxed{ \lfloor \sqrt{n} \rfloor } \).