A convex polyhedron has $m$ triangular faces (there can be faces of other kind too). From each vertex there are exactly 4 edges. Find the least possible value of $m$.
1. **Define Variables and Use Euler's Formula:**
   Let \( F \) be the number of faces, \( V \) be the number of vertices, and \( E \) be the number of edges of the convex polyhedron. According to Euler's polyhedron formula, we have:
   \[
   F + V = E + 2
   \]
   Given that each vertex is connected by exactly 4 edges, we can express \( E \) in terms of \( V \):
   \[
   E = \frac{4V}{2} = 2V
   \]
   Substituting \( E = 2V \) into Euler's formula, we get:
   \[
   F + V = 2V + 2 \implies F = V + 2
   \]

2. **Sum of the Edges of Faces:**
   Denote \( e_i \) as the number of edges of the face \( f_i \) for \( 1 \le i \le F \). Since each edge belongs to exactly 2 faces, the sum of the edges of all faces is:
   \[
   e_1 + e_2 + \dots + e_F = 2E
   \]
   For triangular faces, \( e_k = 3 \), and for other faces, \( e_j \ge 4 \). Let \( m \) be the number of triangular faces. Then, the sum of the edges can be expressed as:
   \[
   2E = 3m + \sum_{j \neq k} e_j
   \]
   Since \( e_j \ge 4 \) for non-triangular faces, we have:
   \[
   2E \ge 3m + 4(F - m)
   \]

3. **Substitute and Simplify:**
   Substitute \( E = 2V \) and \( F = V + 2 \) into the inequality:
   \[
   2(2V) \ge 3m + 4((V + 2) - m)
   \]
   Simplify the inequality:
   \[
   4V \ge 3m + 4V + 8 - 4m
   \]
   \[
   0 \ge 3m - 4m + 8
   \]
   \[
   0 \ge -m + 8
   \]
   \[
   m \ge 8
   \]

Thus, the least possible value of \( m \) is 8.

The final answer is \( \boxed{8} \).