We have the following two observations.

Lemma 1. The surface of any convex polyhedron $D$ can be projected to a plane so that the faces of $D$ are in one-to-one correspondence with the faces of a planar graph $G$ and edges of the $D$ are in one-to-one correspondence with the edges of $G$.

Proof. First, we choose a sufficiently small sphere $H$ within this convex polyhedron and project this polyhedron onto the spherical surface. Then choose a point $P$ on the sphere which is different from the projection images of the vertices of the polyhedron. An inversion in a sphere centered at $P$ turns the sphere $H$ into a plane.

Lemma 2. Suppose that all faces of a planar graph are bounded by a cycle of even length then every cycle is of even length.

Proof. Consider the cycle $S$ of the planar graph $G$. Notice that we can subdivide the cycle $S$ into disjoint cycles $S_1, S_2, \dots, S_n$, where each cycle $S_i$ cannot be divided into smaller cycles. Consider any cycle $S_i$, the region bounded by $S_i$ can be partitioned into disjoint union of faces of $G$. This implies that each cycle $S_i$ has even length since each face of $G$ has even length. Therefore the cycle $S$ has even length.

Let $G$ be the planar graph obtained from the polyhedron $D$ by using Lemma 1. We denote its dual graph by $G'$. The vertices of $G'$ are the faces of $G$, and two vertices are adjacent iff the corresponding two faces have a common edge in $G$. This dual graph $G'$ is also a planar graph. The faces of the planar graph $G'$ correspond to the vertices of the graph $G$. Because every vertex of $G$ has even degree, every face of $G'$ is bounded by even cycles. By Lemma 2, $G'$ has no odd cycles, it is bipartite. Therefore, the vertices of $G'$ can be colored in red and blue such that no two adjacent vertices have the same color. In other words, the faces of $G$ can be colored in red and blue such that no two faces which share a common edge are of the same color. Combining this with Lemma 1, we see that the faces of the convex polyhedron $D$ can be colored in red and blue such that two faces that share an edge have different colors. Without loss of generality, suppose that $F$ is colored blue.

We denote by $S_A$ the sum of the numbers assigned on the edges of the face $A$. We have
$$S_F = \sum_{A \text{ is a red face}} S_A - \sum_{A \text{ is a blue face, } A \neq F} S_A.$$
By the hypothesis, $S_A$ is divisible by 2024 for all the faces $A \neq F$. So each term on the right hand side of the above equality is divisible by 2024. We deduce that $S_F$ is divisible by 2024.

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