$12 \cdot 100$ For which $n$, does there exist a polyhedron with $n$ edges?



Translating the text into English, the result is as follows:

$12 \cdot 100$ For which $n$, does there exist a polyhedron with $n$ edges?
［Solution］(1) Since each polyhedron has at least 4 vertices, and each vertex leads to at least 3 edges, each polyhedron must have at least 6 edges. A tetrahedron has exactly 6 edges.
(2) A polyhedron with 7 edges does not exist. If such a polyhedron existed, when it has a face that is a polygon with at least 4 sides, each vertex on this face would lead to at least 1 edge outside the face, thus the polyhedron would have at least 8 edges. Therefore, each face of the polyhedron must be a triangle. Hence, the number of edges of the polyhedron is \( k = \frac{3}{2} S \), where \( S \) is the number of faces of the polyhedron. This implies that \( k \) must be a multiple of 3, and thus cannot be 7, leading to a contradiction.
(3) When \( m \geqslant 4 \), a pyramid with an \( m \)-sided base is a polyhedron with \( 2m \) edges.
(4) Starting from a pyramid with \( k = 2(m-1) \) edges, cutting off one corner of the base can result in a polyhedron with \( n = 2m + 1 \) edges (\( m \geqslant 4 \)). In fact, let \( M, N, P \) be the midpoints of the 3 edges extending from the vertex \( S \) of the pyramid's base. By making a plane through these 3 points and cutting off the tetrahedron \( SMNP \), the resulting polyhedron will have exactly \( 2m + 1 \) edges.

In summary, a polyhedron with \( n \) edges exists if and only if \( n \geqslant 6 \) and \( n \neq 7 \).