Let $n \ge 3$ be a fixed integer.  The numbers $1,2,3, \cdots , n$ are written on a board.  In every move one chooses two numbers and replaces them by their arithmetic mean.  This is done until only a single number remains on the board.

Determine the least integer that can be reached at the end by an appropriate sequence of moves.

(Theresia Eisenkölbl)
1. **Initial Setup**: We start with the numbers \(1, 2, 3, \ldots, n\) on the board.

2. **Understanding the Move**: In each move, we choose two numbers \(a\) and \(b\) and replace them with their arithmetic mean \(\frac{a+b}{2}\).

3. **Sum of Numbers**: The sum of the numbers on the board initially is:
   \[
   S = 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2}
   \]
   This is the sum of the first \(n\) natural numbers.

4. **Number of Moves**: Since we start with \(n\) numbers and reduce the count by one in each move, we will perform \(n-1\) moves to end up with a single number.

5. **Invariant Property**: The sum of the numbers on the board is halved in each move. However, the sum of the numbers on the board remains constant throughout the process because replacing \(a\) and \(b\) with \(\frac{a+b}{2}\) does not change the total sum.

6. **Final Number**: After \(n-1\) moves, the single remaining number must be the average of the initial sum of the numbers. Therefore, the final number is:
   \[
   \frac{S}{n} = \frac{\frac{n(n+1)}{2}}{n} = \frac{n+1}{2}
   \]

7. **Least Integer**: The least integer that can be reached is the smallest integer greater than or equal to \(\frac{n+1}{2}\). For \(n \ge 3\), this value is:
   \[
   \left\lceil \frac{n+1}{2} \right\rceil
   \]

8. **Verification**: For \(n = 3\), the numbers are \(1, 2, 3\). The sum is \(6\), and the average is \(2\). For \(n = 4\), the numbers are \(1, 2, 3, 4\). The sum is \(10\), and the average is \(2.5\), which rounds up to \(3\). For \(n = 5\), the numbers are \(1, 2, 3, 4, 5\). The sum is \(15\), and the average is \(3\).

Thus, the least integer that can be reached is:
\[
\left\lceil \frac{n+1}{2} \right\rceil
\]

The final answer is \( \boxed{ \left\lceil \frac{n+1}{2} \right\rceil } \).