Example 1 A square piece of paper, cut along a straight line that does not pass through any vertex to divide it into two parts; take out one of the parts, and cut it along a straight line that does not pass through any vertex to divide it into two parts; then from the three parts obtained, take out one of them, and still cut it along a straight line that does not pass through any vertex to divide it into two parts... In the end, 34 sixty-two-sided polygons and some other polygonal paper pieces are obtained. Then the minimum number of cuts needed is ( ).
(2006, National Junior High School Mathematics Competition)
(A) 2004
(B) 2005
(C) 2006
(D) 2007
According to the problem, when the front knife cuts along a straight line that does not pass through a vertex, dividing it into two parts, each cut increases the sum of the interior angles of each part by $360^{\circ}$. Thus, after $k$ cuts, there are $k+1$ polygons, and the sum of the interior angles of these polygons is $(k+1) \times 360^{\circ}$.

Since there are 34 hexacontakaihexagons (62-sided polygons) among these $k+1$ polygons, their sum of interior angles is
$$
34 \times (62-2) \times 180^{\circ} = 34 \times 60 \times 180^{\circ},
$$

and the remaining polygons number $(k+1) - 34 = k - 33$.
The sum of the interior angles of these polygons is at least $(k-33) \times 180^{\circ}$, so
$$
\begin{array}{l}
(k+1) \times 360^{\circ} \\
\geqslant 34 \times 60 \times 180^{\circ} + (k-33) \times 180^{\circ}.
\end{array}
$$

Solving this, we get $k \geqslant 2005$.
By making 2005 cuts in the following manner, the required result can be achieved.

First, cut off 1 triangle from a square, resulting in 1 triangle and 1 pentagon; then cut off 1 triangle from the pentagon, resulting in 2 triangles and 1 hexagon... After 58 cuts, 58 triangles and 1 hexacontakaihexagon (62-sided polygon) are obtained. Next, take 33 triangles and cut each one, resulting in 33 triangles and 33 quadrilaterals. For these 33 quadrilaterals, use the same method as for the square, and cut each one 58 times to get 34 hexacontakaihexagons and $33 \times 58$ triangles.
Thus, a total of
$$
58 + 33 + 33 \times 58 = 2005 \text{ (cuts). }
$$

Therefore, the answer is B.