A staircase-brick with 3 steps of width 2 is made of 12 unit cubes. Determine all integers $ n$ for which it is possible to build a cube of side $ n$ using such bricks.
1. **Volume Consideration**:
   - The volume of a staircase-brick with 3 steps of width 2 is \(12\) unit cubes.
   - For a cube of side \(n\) to be built using these bricks, its volume \(n^3\) must be a multiple of \(12\).
   - Therefore, \(n^3 \equiv 0 \pmod{12}\).

2. **Divisibility by 6**:
   - Since \(12 = 2^2 \times 3\), \(n^3\) being a multiple of \(12\) implies \(n\) must be a multiple of \(6\).
   - Thus, \(n = 6k\) for some integer \(k\).

3. **Constructibility for Multiples of 12**:
   - We claim that \(n\) can be any multiple of \(12\). 
   - To support this, note that two staircases can be combined to create a \(2 \times 3 \times 4\) box.
   - These \(2 \times 3 \times 4\) boxes can be stacked to fill any cube with side length a multiple of \(12\).

4. **Impossibility for \(n = 12k - 6\)**:
   - Suppose \(n = 12k - 6\) for some positive integer \(k\).
   - We need to show that it is impossible to fill such a cube completely with staircases.

5. **Labeling Subcubes**:
   - Break up the \(n = 12k - 6\) cube into unit subcubes and let their centers be \((a, b, c)\) for \(1 \le a, b, c \le n\).
   - Assign each subcube a label from \(0\) to \(3\) equivalent to the sum of its coordinates modulo \(4\).

6. **Odd and Even Counts**:
   - Each staircase-brick has an odd number of unit cubes of each type (0, 1, 2, 3 modulo 4).
   - 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.

7. **Breaking Down the Cube**:
   - Break up the \(12k - 6\) cube into:
     - One \(12k - 4\) cube,
     - Three \((12k - 4) \times (12k - 4) \times 2\) prisms,
     - Three \((12k - 4) \times 2 \times 2\) prisms,
     - 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.
   - The \(2 \times 2 \times 2\) cube has an odd number of unit cubes of each type.

8. **Contradiction**:
   - This means that the big cube has an odd number of unit cubes of each type, which contradicts the requirement that the cube must have an even number of unit cubes of each type.

\(\blacksquare\)

The final answer is \( \boxed{ \text{all multiples of } 12 } \).