Determine all values of $n$ such that it is possible to divide a triangle in $n$ smaller triangles such that there are not three collinear vertices and such that each vertex belongs to the same number of segments.
1. **Identify the problem and the given values**: We need to determine all values of \( n \) such that it is possible to divide a triangle into \( n \) smaller triangles, ensuring no three vertices are collinear and each vertex belongs to the same number of segments.

2. **Graph Theory Representation**: Consider the planar graph formed by the vertices and edges of the triangles. Since no three vertices are collinear, each face of this graph is a triangle. Thus, there are \( f = n \) faces.

3. **Counting Edges**: Each triangular face uses exactly 3 edges, and each edge belongs to exactly 2 triangles, except for the three edges of the initial triangle. Therefore, the total number of edges \( e \) is given by:
   \[
   e = \frac{3n - 3}{2} + 3 = \frac{3(n + 1)}{2}
   \]

4. **Odd \( n \)**: Since \( e \) must be an integer, \( n \) must be odd.

5. **Vertex Degree and Euler's Formula**: Let \( d \) be the common degree of the vertices, and \( v \) be the number of vertices. The sum of the degrees of the vertices is twice the number of edges:
   \[
   dv = 2e = 3(n + 1)
   \]
   Using Euler's formula for planar graphs \( v - e + f = 2 \) (including the unbounded face), we get:
   \[
   v - \frac{3(n + 1)}{2} + n = 2
   \]
   Simplifying, we find:
   \[
   v = \frac{n + 5}{2}
   \]

6. **Degree Calculation**: From the degree sum formula, we have:
   \[
   d \left( \frac{n + 5}{2} \right) = 3(n + 1)
   \]
   Solving for \( d \):
   \[
   d = \frac{6(n + 1)}{n + 5}
   \]

7. **Odd \( n \) Analysis**: Let \( n = 2k + 1 \) where \( k \geq 0 \). The degree \( d \) becomes:
   \[
   d = \frac{6(k + 1)}{k + 3}
   \]
   This must be an integer.

8. **Case 1: \( k \) is even**: If \( k \) is even, \( \gcd(k + 1, k + 3) = 1 \), so \( k + 3 \) must divide 6. Since \( k + 3 \) is odd and greater than 1, the only possibility is \( k = 0 \), giving \( n = 1 \).

9. **Case 2: \( k \) is odd**: Let \( k = 2q + 1 \). Then:
   \[
   d = \frac{6(q + 1)}{q + 2}
   \]
   Since \( \gcd(q + 1, q + 2) = 1 \), \( q + 2 \) must divide 6. The possible values are \( q + 2 = 2, 3, 6 \), leading to \( q = 0, 1, 4 \), and thus \( n = 3, 7, 19 \).

10. **Verification**:
    - \( n = 1 \): The initial triangle itself.
    - \( n = 3 \): Choose an interior point and join it to the vertices.
    - \( n = 7 \): Use Morley's configuration with three interior points.
    - \( n = 19 \): Construct a configuration with 12 vertices, each of degree 5, forming 19 triangles.

The final answer is \( \boxed{ n = 1, 3, 7, 19 } \).