$k=1-$ the task cannot be solved, because three different weights cannot be divided into three equal weight parts. If $k$ is even, that is, $k=2 m$ for a natural number $m$, then put $m$ empty and $m$ full containers on the first and second trucks, and the remainder on the third: all the half-full ones. If $k>1$ and odd, that is, $k=2 m+1$ for a natural number $m$, then put $m$ empty, $m$ full, and $1$ half-full container on the first and second trucks, and the remainder on the third: $1$ empty, $1$ full, and $(k-2)$ half-full containers. It is clear that this way, on average, $k$ half-full containers end up on each truck, so the load is equal.

Consequently, the task cannot be solved for $k=1$, and it is solvable for every integer $k$ greater than one.

Remarks. In the submitted solutions, we read several methods for distributing the containers. We will present some of these ideas.

1. There exists a natural number $m$ such that $k / 4 < m < k / 2$ if $k \geqq 2$. Put $m$ empty and $m$ full containers on the first and second trucks, and $(k-2 m)$ half-full containers, and the remainder on the third.
2. 2 containers can be distributed as follows: one empty and one full on the first and second trucks, and two half-full on the third. For even $k$, repeat this method to distribute the containers. For odd $k$, if $k \geqq 3$, take 3 containers from the total and put one of each type on each truck. The remainder is even, so it can be distributed using the previous method.
3. For every $k \geqq 2$, $k$ can be written in the form $k = 3i + 2j$, where $i$ and $j$ are non-negative integers. It is enough to show the even distribution for $k=2$ and $k=3$, and then repeat these $j$ times and $i$ times, respectively.
4. By complete induction: 2 containers can be distributed. If it is distributable for some $k$, then for $(k+1)$: take one half-full container from one truck and replace it with one empty and one full container. Place the remaining two halves on the other two trucks.
3 trucks need to transport $k$ full containers, $k$ half-full containers, and $k$ empty containers in such a way that each truck is equally loaded. For which natural number $k$ can this task be solved?