Example 4 Several 1s and 2s are arranged in a row
$$
1,2,1,2,2,1,2,2,2,1,2, \cdots
$$

The rule is: the 1st number is 1, the 2nd number is 2, the 3rd number is 1, ... Generally, first write a row of 1s, then insert $k$ 2s between the $k$th 1 and the $(k+1)$th 1 ($k=1$, $2, \cdots$). Try to answer:
(1) Is the 2005th number 1 or 2?
Explanation: Clearly, the position of 1 is somewhat special. For the convenience of calculation, we might as well divide this sequence of numbers into $n$ groups: the 1st group has 1 number, the 2nd group has 2 numbers, the 3rd group has 3 numbers, ..., the $n$th group has $n$ numbers, and the last number of each group is 1, while the rest are 2, i.e.,
(1), $(2,1),(2,2,1), \cdots,(\underbrace{2,2, \cdots, 2,1}_{n-1})$.
Thus, to determine whether the 2005th number is 1 or 2, we just need to use the Gauss summation formula to find out which group the 2005th number is in.

Below, we will use the trial method to estimate.
These $n$ groups have a total of $1+2+\cdots+n=\frac{n(n+1)}{2}$ numbers.
When $n=62$, there are 1953 numbers;
When $n=63$, there are 2016 numbers.
From this, we can see that the 2005th number is in the 63rd group, and it is not the last number.
Therefore, the 2005th number is 2.