Given any triangle $ABC$ and any positive integer $n$, we say that $n$ is a [i]decomposable[/i] number for triangle $ABC$ if there exists a decomposition of the triangle $ABC$ into $n$ subtriangles with each subtriangle similar to  $\triangle ABC$. Determine the positive integers that are decomposable numbers for every triangle.
1. **Understanding the Problem:**
   We need to determine the positive integers \( n \) such that any triangle \( \triangle ABC \) can be decomposed into \( n \) subtriangles, each similar to \( \triangle ABC \).

2. **Initial Consideration:**
   Let's consider a triangle \( \triangle ABC \) with side length 1. For a number \( n \) to be decomposable, we need to find a way to divide \( \triangle ABC \) into \( n \) smaller triangles, each similar to \( \triangle ABC \).

3. **Basic Decomposition:**
   - For \( n = 1 \), the triangle is itself, so it is trivially decomposable.
   - For \( n = 2 \), it is not possible to decompose a triangle into 2 smaller triangles that are similar to the original triangle.

4. **Decomposition for \( n = 4 \):**
   - We can decompose \( \triangle ABC \) into 4 smaller triangles, each similar to \( \triangle ABC \). This can be done by drawing lines parallel to the sides of \( \triangle ABC \) at one-third of the height from each vertex, creating 4 smaller triangles.

5. **Generalizing the Decomposition:**
   - From \( n = 4 \), we can generate more decomposable numbers by subdividing one of the smaller triangles into 4 even smaller triangles. This gives us \( n = 4, 7, 10, 13, \ldots \) (i.e., \( n = 4 + 3k \) for \( k \geq 0 \)).
   - Similarly, starting from \( n = 6 \), we can generate \( n = 6, 9, 12, 15, \ldots \) (i.e., \( n = 6 + 3k \) for \( k \geq 0 \)).
   - Starting from \( n = 8 \), we get \( n = 8, 11, 14, 17, \ldots \) (i.e., \( n = 8 + 3k \) for \( k \geq 0 \)).

6. **Non-decomposable Numbers:**
   - The only missing values are \( n = 2, 3, 5 \). It is easy to see that these numbers are not feasible because there is no way to decompose a triangle into 2, 3, or 5 smaller triangles that are similar to the original triangle.

Conclusion:
The positive integers that are decomposable numbers for every triangle are those of the form \( n = 1 \) or \( n = 3k + 1 \) for \( k \geq 1 \).

The final answer is \( \boxed{ n = 1 } \) or \( n = 3k + 1 \) for \( k \geq 1 \).