Let's analyze the process of elimination step by step, starting from the list of integers from 1 to 2002:

### Step 1:
Initially, the numbers $1, 2, \ldots, 2002$ are written on the board. In this first step, numbers at positions $1, 4, 7, \ldots$ (i.e., $(3k+1)$-th positions for $k = 0, 1, 2, \ldots$) are erased.

These positions correspond to the arithmetic sequence:
$$3k + 1 \quad (k = 0, 1, 2, \ldots)$$

### Step 2:
After the first round of erasure, the sequence is reduced, starting from 2 with a common difference of 3:
$$2, 3, 5, 6, 8, 9, \ldots$$

Now, we erase numbers again at the positions $(3k+1)$-th of the new sequence, specifically items such as 2, 5, 8, $\ldots$

### Subsequent Steps:
This process continues iteratively. After each round of erasure, we get a new list of numbers. The sequence of positions $3k + 1$ will become sparser, and the number of remaining numbers will decrease.

### Key Observation:
Because we have to erase at positions $3k+1$ continually and remove the numbers in those positions, and given the dense initial set from 1 to 2002, every remaining number after a complete iteration follows the form $2, 6, 10, \ldots$, which translates to:

$$2 + 4n \quad (n = 0, 1, 2, \ldots)$$

Each sequence formed here is 4 positions apart.

### Final Step:
This removal efficiently sequences down the integers until only one number remains. The pattern observed means the result can only be consistent with the arithmetic progression defined by terms like $2, 6, 10, \ldots$.

### Conclusion:
As such numbers maintain their presence down every possible iteration due to the arithmetic progression nature and group nature (distance multiplying by a factor each time), these terms remain consistent choices. Since after enough eliminations these sequences essentially divide down until a single sequence element sustains:

Thus, the number remaining is:
$$\boxed{2,6,10}$$
This step-based approach confirms that numbers fitting this gap are never fully eliminated in any active $3k+1$ sequence defined reductions, but maintaining membership in all-remaining $2,6,10,\ldots$ ones.