Example 13 In the interior of a square piece of paper, 2005 points are given. These points and the 4 vertices of the square form a set $M$, where no three points are collinear. Then, the square paper is cut into some triangles according to the following rules: (1) each vertex of the triangles is a point in $M$; (2) except for the vertices, no triangle contains any point in $M$. How many triangles can be cut out in total? How many cuts are needed in total? (One cut is needed to cut out one side of a triangle.)
Let the total number of triangles cut out be $x$, and the total number of cuts required be $y$. Then, the sum of the interior angles of these $x$ triangles is $180^{\circ} x$. On the other hand, at each of the 2005 points inside the square, the sum of the interior angles of the triangles is $360^{\circ}$, and at each vertex of the square, the sum of the interior angles of the triangles is $90^{\circ}$. Therefore, the total sum of the interior angles of all the triangles is $2005 \times 360^{\circ} + 4 \times 90^{\circ}$. Thus, $180^{\circ} x = 2005 \times 360^{\circ} + 4 \times 90^{\circ}$. Solving this, we get $x = 4012$ (triangles).

These triangles have a total of $4012 \times 3$ sides. On the other hand, each previous cut produces two sides of the triangles (the common side of two triangles), plus the 4 sides of the square, so the total number of sides of these triangles is also $2 y + 4$. Therefore, $2 y + 4 = 4012 \times 3$. Solving this, we get $y = 6016$ (cuts).