Since the data of the problem concern $n$ real numbers $x_1, x_2, \dots, x_n$ in the closed interval $[-4, 2]$, we consider the polynomial
$$P(x) = (x+4)(x-2)(x-1)^2,$$
which in $[-4, 2]$ satisfies the relation
$$P(x) = (x+4)(x-2)(x-1)^2 \le 0. \quad (1)$$
Adding by parts the inequalities coming from (1) for $x = x_1, x_2, \dots, x_n$, and taking in mind the conditions of the problem, we find:
$$0 \ge P(x_1) + \dots + P(x_n) = \sum_{i=1}^{n} x_i^4 - 11 \sum_{i=1}^{n} x_i^2 + 18 \sum_{i=1}^{n} x_i - 8n \ge 34n - 11 \cdot 4n + 18n - 8n = 0. \quad (2)$$
Hence, since $P(x_i) \le 0$, for all $i = 1, 2, \dots, n$, from relation (2) we have: $P(x_i) = 0$, for all $i = 1, 2, \dots, n$, which means that $x_i \in \{-4, 1, 2\}$, for all $i = 1, 2, \dots, n$. We suppose that from the integers $x_1, x_2, \dots, x_n$, $a$ are equal to $-4$, $b$ are equal to $1$ and $c$ are equal to $2$. Then we have $a+b+c=n$ and from the data of the problems we have the inequalities
$$\left\{ \begin{array}{l} -4a+b+2c \ge a+b+c \\ 16a+b+4c \le 4(a+b+c) \\ 256a+b+16c \ge 34(a+b+c) \end{array} \right\} \Leftrightarrow \left\{ \begin{array}{l} c \ge 5a \\ b \ge 4a \\ 222a \ge 33b+18c \end{array} \right\}.$$
By multiplying both parts of the first inequality with $18$ and the second with $33$ and summing the produced inequalities by parts we get the inequality
$$33b+18c \geq 132a+90a = 222a,$$
which in combination with the inequality $222a \geq 33b+18c$ gives:
$$33b+18c = 222a,$$
which is valid, if and only if $b = 4a$, $c = 5a$, that is
$$a+b+c=10a \text{ or } 10a=n.$$
Therefore the numbers $x_1, x_2, \dots, x_n$ there exist, if and only if, $n$ is a multiple of $10$. For $n=10m$, where $m$ is a positive integer a possible solution arises by taking $m$ times the number $-4$, $4m$ times the number $1$ and $5m$ times the number $2$.