Six. (15 points) A square is divided into 4 squares, the number of edges in the division diagram is 12. If a square is divided into 2005 convex polygons, try to find the maximum number of edges in the division diagram.


Six, according to Euler's theorem, the number of vertices $a$, the number of faces $b$, and the number of edges $e$ of a simple polyhedron have the following relationship: $a+b-e=2$.

From Euler's theorem, it is easy to see that if a convex polygon is divided into $n$ convex polygons, then the number of vertices $a$, the number of polygons $n$, and the number of edges $e$ in the division graph satisfy
$$
a+n-e=1 \text {. }
$$

Below, in the general case, that is, when a square is divided into $n$ convex polygons, find the maximum value of the number of edges in the division graph. Let the number of vertices in the division graph be $a$, the number of polygons be $n$, and the number of edges be $e$.
(1) First, find the upper bound of the number of edges.
Let the four vertices of the original square be $A, B, C, D$. If the vertex $V \notin\{A, B, C, D\}$ of the convex polygon, it is easy to know that
$d(V) \geqslant 3$ (here $d(V)$ represents the number of edges passing through vertex $V$). Hence, $d(V) \leqslant 3(d(V)-2)$.
Noting that there are $a-4$ such vertices $V$, we have $a-4$ inequalities of the above form. Adding them up and noting that each edge, except for the four sides of the square, is the edge of two convex polygons, we have
$$
\begin{array}{l}
2 e-[d(A)+d(B)+d(C)+d(D)] \\
\leqslant 3[2 e-(d(A)+d(B)+d(C)+d(D))]- \\
\quad 6(a-4),
\end{array}
$$

i.e., $4 e \geqslant 2(d(A)+d(B)+d(C)+d(D))+6(a-4)$.
Since $d(A) \geqslant 2, d(B) \geqslant 2, d(C) \geqslant 2, d(D) \geqslant 2$,
then $2 e \geqslant 8+3(a-4)=3 a-4$.
From equation (1), we have
$$
3 a+3 n-3 e=3 \text {. }
$$

Substituting equation (2) into equation (3) and rearranging, we get
$$
3 e+3=3 a+3 n \leqslant 2 e+4+3 n,
$$

i.e., $e \leqslant 3 n+1$.
(2) Construct an example where the number of edges $e=3 n+1$.
As shown in Figure 3, draw $n-1$ parallel lines to one side of the square successively, dividing the square into $n$ rectangles. It is easy to see that the number of edges
$$
\begin{array}{l}
e=4+3(n-1) \\
=3 n+1 .
\end{array}
$$

In summary, the maximum value of the number of edges in the division graph is $3 n+1$.
Therefore, when the square is divided into 2005 convex polygons, the maximum number of edges is 6016.
(Wang Xiaohui provided)