Example 1: A person places 2013 balls in sequence on a circle according to the following rules: First, place the first ball, then move counterclockwise around the circle, and whenever passing a previously placed ball, place a new ball in the gap; continue this process. The ball placed on the $k$-th time is numbered $k$. After all the balls are placed, find the numbers of the two balls adjacent to the 2013rd ball.
Solve by the method of unraveling step by step, considering layers.

When the 1st and 2nd balls are placed, continue around one circle, then place the 3rd and 4th balls; after the second circle, place the 5th, 6th, 7th, and 8th balls; after the third circle, place the 9th, 10th, ..., 16th balls; ... after the k-th circle, place the \(2^k + 1, 2^k + 2, \cdots, 2^{k+1}\) balls.

Now consider the k-th circle, call the \(2^k\) balls placed in the k-th circle "new balls," and all the \(2^k\) balls placed in previous circles "old balls."

Then, after the k-th circle's balls are placed, each new ball must be surrounded by old balls, and each old ball must be surrounded by new balls. The \(2^k\) balls placed in the k-th circle should be inserted between the \(2^{k-1}\) balls placed in the (k-1)-th circle.

Therefore, for the m-th ball placed in the (k-1)-th circle, the ball inserted in front (in the direction of movement) is the (2m)th ball placed in the k-th circle, and the ball inserted behind is the (2m-1)th ball.

Thus, for the n-th ball placed in the k-th circle, if n is even, then the ball behind it is the ball placed in the (k-1)-th circle, with the number \(\frac{n}{2}\); if n is odd, then the ball in front of it is the ball placed in the (k-1)-th circle, with the number \(\frac{n+1}{2}\).

Since \(2^{10} < 2013 < 2^{11}\), the 2013th ball is placed in the 10th round.

Because 2013 is odd, the ball in front of it is the ball placed in the 9th round, with the number \(\frac{2013+1}{2} = 1007\).
Next, calculate the number of the ball behind the 2013th ball (let it be \(k\)).
Remove the balls placed in the 10th circle, then the ball \(k\) behind the 2013th ball becomes the ball behind the 1007th ball placed in the 9th round.

Since 1007 is also odd, the ball in front of it is the ball placed in the 8th round, with the number \(\frac{1007+1}{2} = 504\). Remove the balls placed in the 9th circle, then the ball \(k\) behind the 2013th ball becomes the ball behind the 504th ball placed in the 8th round.

Since 504 is even, the ball behind it has the number \(\frac{504}{2} = 252\), i.e., \(k = 252\).

Therefore, the ball in front of the 2013th ball is the 1007th ball, and the ball behind it is the 252nd ball.