We will first prove that coverings can be found if $n = 3k$ and if $n = 3k - 1$. When $k = 1$ these are the cases $n = 2$ and $n = 3$ for which the diagrams below show a covering by trexes. In these diagrams, the unit hexagons are represented by their centres.

To extend this for all $k \ge 1$, we observe that, for any $n \ge 1$, the region obtained by removing an $n$-level honeycomb from an $(n+3)$-level honeycomb can be covered by trexes without overlap, see left diagram below. The statement follows now from the cases $n = 2$ and $n = 3$.

Alternatively, we may cover the large hexagon formed by the centres of the unit hexagons by $3$ congruent parallelograms, each covering $n(n+1)$ centres, as shown in the diagram on the right below.

When the number of hexagon centres along one side of such a parallelogram is divisible by $3$, we can cover it by trexes. More specifically:

* if $n = 3k$ then each parallelogram of size $3k(3k+1)$ can be covered by $k(3k+1)$ trexes;
* if $n = 3k + 2$ then each parallelogram of size $(3k+2)(3k+3)$ can be covered by $(k+1)(3k+2)$ trexes.

Next we prove that no coverings exist if $n = 3k+1$. We present three different proofs. In all of them we label the centres of the hexagons by $0$, $1$, $2$ with $1$ in the centre of the honeycomb and so that two hexagons with the same label never share an edge. It follows that every trex contains each label $0$, $1$, $2$ exactly once, hence the honeycomb (with central hexagon removed) can only be covered with trexes when it contains the same number of $0$-s, $1$-s and $2$-s.

We now prove that this is not the case when $n = 3k + 1$.

Proof 1. We note that the labelling is preserved by rotation around the centre with angle $120^\circ$. Hence by splitting the honeycomb minus the centre into three parallelograms of sides $n = 3k + 1$ and $n + 1 = 3k + 2$ as above, we note that the numbers of $0$-s, $1$-s and $2$-s in each parallelogram must be one third of the total numbers of $0$-s, $1$-s and $2$-s in the honeycomb minus the centre. However, the number of hexagon centres in such a parallelogram is $(3k + 1)(3k + 2)$ which is not a multiple of $3$ and so cannot contain the same number of $0$-s, $1$-s and $2$-s. Hence the numbers of $0$-s, $1$-s and $2$-s in the honeycomb minus the centre cannot be equal to each other.

Proof 2. As we have seen above, the region obtained by removing an $n$-level honeycomb from an $(n+3)$-level honeycomb can be covered by trexes without overlap. This implies that the numbers of $0$-s, $1$-s and $2$-s in an $(n+3)$-level honeycomb (with centre removed) coincide if and only if they do so in an $n$-level honeycomb (with centre removed).
Because in case $n = 1$ there are three hexagons labelled $0$ but no hexagon labelled $1$, the numbers of $0$-s and $1$-s in a $(3k+1)$-level honeycomb (with centre removed) do not coincide.

Proof 3. For each $a \in \{0, 1, 2\}$ and each $n$, let $h_n(a)$ denote the number of hexagons labelled $a$ in $H_n$, the $n$-level honeycomb (including the central hexagon). We wish to find $h_n(a) - h_0(a)$. Let
$$c_n(a) = h_n(a) - h_{n-1}(a) = b_n(a) + v_n(a)$$
denote the number of labels $a$ on the collar $H_n - H_{n-1}$, where $v_n(a)$ counts the labels at the $6$ corners and $b_n(a)$ counts the remaining vertices.

We can compare $H_n - H_{n-1}$ with $H_{n-2} - H_{n-3}$, as the labels at the outer border (except at the corner) repeat the labels from the inner border. Hence the following recurrences for $n \ge 3$:
$$b_n(a) = b_{n-2}(a) + 2v_{n-2}(a) \quad \text{and so}$$
$$c_n(a) = b_n(a) + v_n(a) = c_{n-2}(a) + v_{n-2}(a) + v_n(a)$$
while $v_3(1) = 6$ and for $n \ge 4$ we have $v_n(a) = v_{n-3}(a)$. Also using $h_n(a) = h_{n-1}(a) + c_n(a)$ we can fill in the following table.

Hence after removing the central hexagon, we have shown that the $n$-level honeycomb cannot be covered by trexes if $n = 3k+1$, since at least $3$ hexagons labelled $1$ would remain uncovered.

Final answer:

All $n$ except those congruent to $1$ modulo $3$ (i.e., all $n$ such that $n \not\equiv 1 \pmod{3}$) can be covered by trexes as described.