Determine all integers $n\ge1$ for which there exists $n$ real numbers $x_1,\ldots,x_n$ in the closed interval $[-4,2]$ such that the following three conditions are fulfilled:
- the sum of these real numbers is at least $n$.
- the sum of their squares is at most $4n$.
- the sum of their fourth powers is at least $34n$.
[i](Proposed by Gerhard Woeginger, Austria)[/i]
To determine all integers \( n \ge 1 \) for which there exist \( n \) real numbers \( x_1, \ldots, x_n \) in the closed interval \([-4, 2]\) such that the given conditions are fulfilled, we will analyze the constraints step by step.

1. **Sum of the numbers:**
   \[
   \sum_{i=1}^n x_i \ge n
   \]

2. **Sum of the squares:**
   \[
   \sum_{i=1}^n x_i^2 \le 4n
   \]

3. **Sum of the fourth powers:**
   \[
   \sum_{i=1}^n x_i^4 \ge 34n
   \]

We start by considering the polynomial \(\sum_{i=1}^n (x_i + 4)(x_i - 2)(x_i - 1)^2\). Since \(x_i \in [-4, 2]\), each term \((x_i + 4)(x_i - 2)(x_i - 1)^2 \le 0\). Therefore, the sum must be nonpositive:
\[
\sum_{i=1}^n (x_i + 4)(x_i - 2)(x_i - 1)^2 \le 0
\]

Expanding and simplifying the polynomial:
\[
(x_i + 4)(x_i - 2)(x_i - 1)^2 = (x_i + 4)(x_i^2 - 4x_i + 4) = x_i^3 - 4x_i^2 + 4x_i + 4x_i^2 - 16x_i + 16 = x_i^3 - 12x_i + 16
\]

Thus:
\[
\sum_{i=1}^n (x_i^3 - 12x_i + 16) \le 0
\]

This implies:
\[
\sum_{i=1}^n x_i^3 - 12 \sum_{i=1}^n x_i + 16n \le 0
\]

Given the conditions, we substitute the constraints:
\[
\sum_{i=1}^n x_i = n, \quad \sum_{i=1}^n x_i^2 = 4n, \quad \sum_{i=1}^n x_i^4 = 34n
\]

We now solve the system of equations for \(x_i \in \{-4, 1, 2\}\). Let \(p\) be the number of terms equal to \(-4\), \(q\) be the number of terms equal to \(1\), and \(r\) be the number of terms equal to \(2\). Then:
\[
p + q + r = n
\]

From the sum of the numbers:
\[
-4p + q + 2r = n
\]

From the sum of the squares:
\[
16p + q + 4r = 4n
\]

From the sum of the fourth powers:
\[
256p + q + 16r = 34n
\]

We solve this system of linear equations. First, we simplify the equations:
\[
-4p + q + 2r = n \quad \text{(1)}
\]
\[
16p + q + 4r = 4n \quad \text{(2)}
\]
\[
256p + q + 16r = 34n \quad \text{(3)}
\]

Subtract equation (1) from equation (2):
\[
(16p + q + 4r) - (-4p + q + 2r) = 4n - n
\]
\[
20p + 2r = 3n
\]
\[
10p + r = \frac{3n}{2} \quad \text{(4)}
\]

Subtract equation (2) from equation (3):
\[
(256p + q + 16r) - (16p + q + 4r) = 34n - 4n
\]
\[
240p + 12r = 30n
\]
\[
20p + r = \frac{5n}{2} \quad \text{(5)}
\]

Subtract equation (4) from equation (5):
\[
\frac{5n}{2} - \frac{3n}{2} = 10p + r - (10p + r)
\]
\[
n = 10p
\]
\[
p = \frac{n}{10}
\]

Substitute \(p = \frac{n}{10}\) into equation (4):
\[
10 \left(\frac{n}{10}\right) + r = \frac{3n}{2}
\]
\[
r = \frac{n}{2}
\]

Finally, substitute \(p = \frac{n}{10}\) and \(r = \frac{n}{2}\) into \(p + q + r = n\):
\[
\frac{n}{10} + q + \frac{n}{2} = n
\]
\[
q = n - \frac{n}{10} - \frac{n}{2}
\]
\[
q = n - \frac{n}{10} - \frac{5n}{10}
\]
\[
q = n - \frac{6n}{10}
\]
\[
q = \frac{4n}{10}
\]
\[
q = \frac{2n}{5}
\]

Thus, \(p = \frac{n}{10}\), \(q = \frac{2n}{5}\), and \(r = \frac{n}{2}\). For these to be integers, \(n\) must be a multiple of 10.

The final answer is \( \boxed{ n } \) must be a multiple of 10.