Find all numbers $n \geq 4$ such that there exists a convex polyhedron with exactly $n$ faces, whose all faces are right-angled triangles.
(Note that the angle between any pair of adjacent faces in a convex polyhedron is less than $180^\circ$.)

[i]Proposed by Hesam Rajabzadeh[/i]
To solve the problem, we need to find all numbers \( n \geq 4 \) such that there exists a convex polyhedron with exactly \( n \) faces, where all faces are right-angled triangles.

1. **Understanding the Problem:**
   - A convex polyhedron is a three-dimensional solid where all interior angles between adjacent faces are less than \( 180^\circ \).
   - Each face of the polyhedron is a right-angled triangle.
   - We need to determine the possible values of \( n \) (the number of faces) for such a polyhedron.

2. **Initial Observations:**
   - Since each face is a triangle, the polyhedron must be composed entirely of triangular faces.
   - The sum of the face angles around each vertex must be less than \( 360^\circ \) for the polyhedron to be convex.

3. **Euler's Formula:**
   - For any convex polyhedron, Euler's formula states:
     \[
     V - E + F = 2
     \]
     where \( V \) is the number of vertices, \( E \) is the number of edges, and \( F \) is the number of faces.

4. **Properties of Right-Angled Triangles:**
   - Each right-angled triangle has one \( 90^\circ \) angle and two \( 45^\circ \) angles.
   - In a polyhedron made of right-angled triangles, the angles around each vertex must sum to less than \( 360^\circ \).

5. **Considering Even \( n \):**
   - We need to show that \( n \) must be even.
   - Each face is a triangle, so each face contributes 3 edges, but each edge is shared by 2 faces. Therefore, the total number of edges \( E \) is:
     \[
     E = \frac{3F}{2}
     \]
   - Substituting \( F = n \) into Euler's formula:
     \[
     V - \frac{3n}{2} + n = 2
     \]
     Simplifying, we get:
     \[
     V + \frac{n}{2} = 2
     \]
     \[
     V = 2 - \frac{n}{2}
     \]
   - For \( V \) to be an integer, \( \frac{n}{2} \) must be an integer, implying \( n \) must be even.

6. **Constructing Examples for Even \( n \):**
   - For \( n = 4 \), consider a tetrahedron. A tetrahedron has 4 faces, each of which can be a right-angled triangle.
   - For \( n = 6 \), consider a cube. A cube can be divided into 6 right-angled triangular faces by cutting along the diagonals of each face.
   - For \( n = 8 \), consider a polyhedron formed by joining two tetrahedra along a common face, resulting in 8 right-angled triangular faces.

7. **Generalization:**
   - By induction, we can construct polyhedra with any even number of faces \( n \geq 4 \) by appropriately joining smaller polyhedra with right-angled triangular faces.

Conclusion:
Thus, the numbers \( n \geq 4 \) for which there exists a convex polyhedron with exactly \( n \) faces, all of which are right-angled triangles, are all even numbers.

The final answer is \( \boxed{ n } \) must be an even number.