Example 6: On the table, there are 1989 same-sized circular paper pieces placed without overlapping. How many different colors are needed at least to color each circular paper piece so that, no matter how these paper pieces are placed, it is always possible to color them in such a way that any two touching circular paper pieces are colored differently?

---

The translation maintains the original text's formatting and structure.
The minimum number of different colors required is 4.
3 colors are not enough. As shown in the figure, if 3 colors are represented by the numbers 1, 2, and 3, circle $A_{1}$ is painted color 1, and $A_{3}$ is painted color 2, then $A_{2}, A_{5}, A_{11}$ can only be painted color 3, and $A_{4}$ can only be painted color 1. Consequently, $A_{5}, A_{8}, A_{10}$ cannot be painted color 3. Suppose they are painted colors 2, 1, and 2, respectively, then $A_{7}, A_{9}$ must both be painted color 3, which leads to a contradiction.

Below is an inductive proof that 4 colors can ensure that we can color $n$ pieces of paper as required.
When $n=1,2,3,4$, it is obviously possible to color them as required.
Assume that when $n=k$, it is possible to color them as required.
When $n=k+1$, take the convex hull of the $k+1$ centers $A_{i} (i=1,2, \cdots, k+1)$, and let $A_{1}$ be one of the vertices of the convex hull. Since the angle at $A_{1}$ is less than $180^{\circ}$, it can be tangent to at most 3 circles $A_{2}, A_{3}, A_{4}$. By the inductive hypothesis, the $k$ circles remaining after removing $A_{1}$ can be colored as required. Since $A_{2}, A_{3}, A_{4}$ use at most 3 colors, painting $A_{1}$ with the 4th color will suffice. Thus, $k+1$ circular pieces of paper can be colored with 4 colors as required.
Therefore, 4 colors can ensure that we can color $n$ pieces of paper as required.
/Comment: Construct a counterexample to prove that 3 colors are not enough; use induction to prove that 4 colors are sufficient. In the second step of the inductive proof, take the convex hull of the centers of $k+1$ circles, then remove one vertex of the convex hull, reducing it to the case of $k$ circles (whose centers may be on the edge or inside the convex hull), which can be handled using the inductive hypothesis.