can someone please find a solution or at least a hint please?

There are 12 unit cubes in each brick, hence $ n$ must be divisible by 2 and 3, i.e., $ 6|n$. Let $ n=6k$. Two staircase-bricks can be used to build a $ 2\times3\times4$ box. Hence it is easy to construct a cube of side $ 12$. Thus $ n=6k$ is possible for even $ k$. We claim that if $ k$ is odd, we cannot build a cube of side $ n$, i.e., the number of bricks must be even. Assume there are $ m$ bricks. Assign them to the Cartesian coordinates such that the center of a corner cube is at (0,0,0). Consider the cubes in $ (x,y,z)$ such that $ x,y,z$ are even. We call such cubes "good". In a brick, there can be either 1 or 3 good cubes. Assume that there are $ p$ bricks that contain 3 good cubes. Hence the total number of good cubes is $ 3p+m-p=2p+m$. But the number of good cubes is $ \frac{12m}{8}$. Hence $ 2p+m=\frac{12m}{8}$, or $ m=4p$. So $ m$ is even. Q.E.D.

Actually you need $ 4|m$ ($ m$ even does not suffice, since there are $ n^3 = 12m$ would just imply $ 8|n^3$, which gives nothing). Or did I miss something?

@above yeah we need 4|m For this we will use comlex combinatorics Let $\epsilon^4=1$ and $\epsilon \neq 1$ and we will assign to each square with coordinates $(x,y,z)$ with $\epsilon^{x+y+z}$ Then for each staircase-brick with starting unit cube $(x,y,z)$ which $x+y+z$ is minimum Then our sum will be 0(it is easy to calculate).And we know that The general sum is $(\epsilon^0+......+\epsilon^{m-1})^3$ and it is 0.Then 4|m and then we are done

@elcinmusazade what is the reasoning for $4|m?$ As I understand it you proved that the sum of the staircase brick is $0$ and the total sum is $0$ and any number of staircase bricks should sum to $0$ so there should be no restriction? jgnr has a good idea and I built on it. Note that $12|n^3$ so $6|n.$ Let $n=6k.$ Note that two staircase bricks can be put together to form a $2\times3\times4$ prism like in the following picture. We can place a three-dimensional array of $6\times4\times3$ of these $2\times3\times4$ prisms to form a $12\times12\times12$ cube. These cubes can then be made into a big cube of side length divisible by $12,$ i.e. when $k$ is divisible by $2.$ Now, we split the $6k\times 6k\times 6k$ into a $3k\times 3k\times 3k$ grid of $2\times 2\times 2$ cubes and in each such cube we call the bottom-left-back corner unit cube special. Note that in the two layers of the staircase brick, one of them cannot have any special cubes. In the above picture, observe that differently-colored cubes cannot be both special, and one of the four colors must be a special cube, so the number of special cubes per staircase brick is $1$ or $3.$ Note that the number of staircase bricks is $18k^3$ which is even. Thus, the number of special cubes is even and so the number of total $2\times 2\times 2$ cubes is even. This implies that $27k^3$ is even, so $k$ is also even. Therefore the only solutions are $n=12k$ for all positive (nonnegative if you happen to count $0$) integers $k$.

Notice that the volume of such a cube must be a multiple of $12$ and thus $n$ should be a multiple of $6$. We claim that $n$ can be $\boxed{\text{all multiples of }12}$. Notice that two staircases can be combined to create a $2\times 3\times 4$ box, which can clearly be stacked to fill any cube with side length a multiple of $12$. It suffices to show that $n=12k-6$ for positive integer $k$ is impossible. Suppose we can fill such a cube completely with staircases. Break up such a cube into unit subcubes and let their centers be $(a, b, c)$ for $1\le a,b,c\le n$. Give each subcube a label from $0$ to $3$ equivalent to the sum of it's coordinates modulo $4$. Now it is easy to check that all staircases placed inside such a cube must have an odd number of unit cubes of every type. Since there are an even number of staircases in filling up a cube, the cube must have an even number of unit cubes of each type. Break up the side length $12k-6$ cube into one $12k-4$ cube, three $(12k-4)\times (12k-4) \times 2$ prisms, three $(12k-4)\times 2\times 2$ prisms, and one $2\times 2\times 2$ cube. Everything except the $2\times 2\times 2$ cube has an even number of unit cubes of each type, and the $2\times 2\times 2$ cube has an odd number of unit cubes of each type. This means that the big cube has an odd number of unit cubes of each type, contradiction.

We claim the answer is all positive multiples of $12$. Evidently, if $n$ is an integer that works, we have $12\mid n^3$, so $6\mid n$. Hence, $n=6k$. Now, note that we may place two staircase-bricks together to form a $2\times3\times4$ prism. Since $2,3,4\mid12$, it readily follows that the cube can be built when $k$ is even. Now we prove the cube cannot be built if $k$ is odd. Suppose it can and consider a tiling of a $6k\times6k\times6k$ cube with staircase-bricks. We separate the brick into six $6\times6$ layers. In each layer, the staircase-brick fills either a $2\times1,2\times2$, or $2\times3$. If the staircase-brick fills a $2\times2$, then it must fill a $2\times1$ and $2\times3$ in the above and below layers. Hence, the bottom-most layer has no $2\times2$ filled by staircase-bricks. It follows the layer right on top is filled only by $2\times2$. It is straightforward to see that the only way to tile a $6k\times6k$ layer using $2\times2$ pieces is to lay out the pieces in a $3k\times3k$ grid. Therefore, we use $9k^2$ pieces. Now, look at the bottom layer. It is filled by $x$ $2\times3$ and $y$ $2\times1$. Since it is the bottom layer, these must be part of one of the $9k^2$ pieces above. Hence, $x+y=9k^2$. On the other hand, summing areas we see $6x+2y=36k^2$, or $3x+y=18k^2$. Thus, $2x=9k^2$. However, $k$ is odd, so this is impossible. Hence, the cube cannot be built when $k$ is odd as desired. $\blacksquare$

The answer is all multiples of $12$. Evidently these work, because we can form $2\times 3\times 4$ bricks by combining two staircase-bricks. Now, for odd multiples of $6$, label each cube with a number from $0$ to $3$, representing its 3-D taxicab distance to some predetermined corner. Note that the number of cubes labeled with each number is odd, since we can "reduce" the dimensions $\bmod\text{ 4}$ to obtain a $2\times 2\times 2$ square, where the counts of all numbers are all odd. Suppose FTSOC that we can form such a cube with staircase-bricks. Then we must clearly use an even number of staircase-bricks. I claim that the counts of each of the labels is odd; this will give a contradiction, since it implies that the counts of each of the numbers in the cube is even. There are two cases to consider. The first case is where the closest cube to the predetermined corner is a cube on the bottom step of the staircase-brick. In this case, each of the labels from $0$ to $3$ appears exactly three times. The second case is where the closest cube is one of the two "back" corners of the staircase-brick. This case gives one label which appears five times, two which appear three times, and one which appears once. Then, we have our desired contradiction, so we are done. $\blacksquare$

We claim that the answer is $12|n$. We can cap the staircase on top of itself to form a 2x3x4 block, which can be combined to build a 12x12x12 block, so all multiples of 12 are possible. Obviously, $n$ is divisible by 6 by volume. Let $n=6m$. Assign coordinates $(i,j,k)$ with $1\leq i,j,k\leq 6m$ to each unit cube. Denote a unit cube as orz if $i,j,k$ are all even. Claim: If a staircase-brick is placed anywhere, it will occupy an odd number of orz unit cubes. Note that the staircase-brick can be decomposed into a 2x2x2 cube and two small 1x1x2 blocks. Note that exactly one unit cube in the 2x2x2 is orz. Now, the four remaining unit cubes can be paired into two pairs such that the two unit cubes in each pair are either both orz or both not orz (since each coordinate differs by either 0 or 2), so there are an odd number of orz unit cubes. Note that the total number of staircase-bricks is $\frac{(6m)^3}{12}=18m^3$, which is even, so the total number of orz cubes must be even since each staircase-brick has an odd number of orz cubes. Therefore, $m$ must be even since there are $27m^3$ total orz cubes, so $n$ must be divisible by 12.

I claim $12|n$ describes all answers. They clearly work as $2,3,4$ all divide $12$ so we can just partition a $12k$ side length cube into small $2\times 3 \times 4 $ rectangular prisms. All numbers that are not multiples of $6$ fail by divisors. Now to show the odd multiples of $6$ fail, FTSOC they work. Then the number of staircases used is $\frac{6^3(2k+1)^3}{24}$ which ends up as odd. This means there is at least one staircase that cannot be paired with another staircase of the same orientation. The reasoning above leads me to think there is no possible tiling, but this is not rigorous. Can someone please give me a push in the right direction, either telling me how to finish from here or that this solution does not work? Thanks.