Example 4: There are 2013 balls placed around a circle, numbered $1, 2, \cdots, 2013$ in a clockwise direction. Starting from a certain ball $a$, and then taking every other ball in a clockwise direction, this process continues until only one ball remains on the circle. For what value of the number of ball $a$ will the last remaining ball be the 2013th ball?
First, consider a special case.
When the ball-picking procedure starts with the ball numbered 1, then when the person picking the balls reaches the next ball to be taken (numbered 3), it is essentially as if the ball numbered 2 has been placed at the farthest point in the direction of his movement. And when he takes the ball numbered 3 and moves to the next ball to be taken (numbered 5), it is again as if the ball numbered 4 has been placed at the farthest point in the direction of his movement. This process continues indefinitely.

Therefore, when the ball-picking procedure starts with the ball numbered 1, the problem is essentially the same as Example 3. In this case, since \(2013 = 2^{10} + 989\), the last remaining ball is numbered \(2 \times 989 = 1978\).
This is not the desired result, but it provides a clue.
If the ball-picking procedure is changed to start from the ball numbered \(x+1\), and the last remaining ball changes from 1978 to 2013, the goal is to find the number of the ball \(a\) by changing the initial position of the ball-picking.

It is easy to see that if the ball numbered \(x+k\) is renumbered as \(k^*\) in a clockwise direction, then the ball originally numbered \(k+1978\) becomes \(1978^*\). Therefore, when the ball-picking procedure starts from \(1^\circ\), the last remaining ball should be numbered \(1978^\circ\).
\[
\begin{array}{l}
\text{Let } 1978^* = 2013 \Rightarrow x + 1978 = 2013 \\
\Rightarrow x = 35 \Rightarrow x + 1 = 36.
\end{array}
\]

Therefore, when the ball \(a\) is numbered 36, the last remaining ball is exactly the ball numbered 2013.