Real numbers are placed at the vertices of an $n{}$-gon. On each side, we write the sum of the numbers on its endpoints. For which $n{}$ is it possible that the numbers on the sides form a permutation of $1, 2, 3,\ldots , n$?

[i]From the folklore[/i]
1. **Case 1: \( n \) is odd**

   Suppose the sides have \(\sigma(1), \sigma(2), \dots, \sigma(n)\) written on them, starting from some vertex. Let \(a\) be the number at the first vertex. Then, the numbers at the vertices can be written as follows:
   - First vertex: \(a\)
   - Second vertex: \(\sigma(1) - a\)
   - Third vertex: \(\sigma(2) - (\sigma(1) - a) = \sigma(2) - \sigma(1) + a\)
   - Fourth vertex: \(\sigma(3) - (\sigma(2) - \sigma(1) + a) = \sigma(3) - \sigma(2) + \sigma(1) - a\)
   - Continue this pattern until the \(n\)-th vertex.

   When we return to the first vertex, we must have:
   \[
   a = \sigma(n) - \sigma(n-1) + \sigma(n-2) - \cdots + (-1)^{n-1} \sigma(1) + (-1)^n a
   \]
   Since \(n\) is odd, \((-1)^n = -1\), so we get:
   \[
   a = \sigma(n) - \sigma(n-1) + \sigma(n-2) - \cdots + (-1)^{n-1} \sigma(1) - a
   \]
   This simplifies to:
   \[
   2a = \sigma(n) - \sigma(n-1) + \sigma(n-2) - \cdots + (-1)^{n-1} \sigma(1)
   \]
   Thus, \(a\) can be determined as:
   \[
   a = \frac{1}{2} \left( \sigma(n) - \sigma(n-1) + \sigma(n-2) - \cdots + (-1)^{n-1} \sigma(1) \right)
   \]
   Therefore, for any permutation \(\sigma\), we can always find a suitable \(a\), and hence it is possible for all odd \(n\).

2. **Case 2: \( n \) is even**

   When \(n\) is even, we obtain a similar system, but now the \(a\)s cancel out. We have:
   \[
   a = \sigma(n) - \sigma(n-1) + \sigma(n-2) - \cdots + (-1)^{n-1} \sigma(1) + (-1)^n a
   \]
   Since \(n\) is even, \((-1)^n = 1\), so we get:
   \[
   a = \sigma(n) - \sigma(n-1) + \sigma(n-2) - \cdots + (-1)^{n-1} \sigma(1) + a
   \]
   This simplifies to:
   \[
   0 = \sigma(n) - \sigma(n-1) + \sigma(n-2) - \cdots + (-1)^{n-1} \sigma(1)
   \]
   Therefore, we must have:
   \[
   \sum_{i=1}^{n} (-1)^i \sigma(i) = 0
   \]
   Working modulo 2, this implies that \(n \equiv 0 \pmod{4}\). Hence, \(n\) must be a multiple of 4.

   For \(n \equiv 0 \pmod{4}\), we can construct a solution by setting:
   \[
   \sigma(4k+1) = 4k+3, \quad \sigma(4k+2) = 4k+4, \quad \sigma(4k+3) = 4k+2, \quad \sigma(4k+4) = 4k+1
   \]
   for all \(0 \leq k < n/4\), and choose \(a = 0\) in the above system. This construction ensures that the sum condition is satisfied.

Therefore, the numbers \(n\) for which this can be done are precisely \(n \not \equiv 2 \pmod{4}\).

The final answer is \( \boxed{ n \not \equiv 2 \pmod{4} } \).