To determine whether it is possible to partition the positive integers into 12 subsets such that for each positive integer $k$, the numbers $k, 2k, \ldots, 12k$ are in different subsets, we will examine the conditions and implications carefully.

First, consider the sequence formed by taking a positive integer $k$ and the multiples $k, 2k, \ldots, 12k$. If these 12 numbers need to be in different subsets, then each multiple must be placed in a unique subset. Therefore, for any set of 12 consecutive multiples starting with any integer $k$, there must be at least 12 subsets.

Assume, for the sake of contradiction, that such a partition exists. Consider the least common multiple of the indices $1, 2, \ldots, 12$, which is 27720. This means that every complete set of multiples $k, 2k, \ldots, 12k$ repeats every 27720 integers.

Now, analyze the numbers $1, 2, \ldots, 12$:
- Each of these 12 numbers must be placed in different subsets to satisfy the condition for $k = 1$.

If we extend this logic to 27720, which covers every complete cycle of divisors up to 12, we see that every sub-cycle of size 12 within a full cycle of 27720 has to find distinct positions in 12 different subsets, as every base divisor and its multiples are unique within a cycle.

With only 12 subsets available, every number $k, 2k, \ldots, 12k$ would ultimately map to a repeating subset selection as $k$ increases, eventually requiring some numbers within a cycle of 27720 to overlap in subsets.

This overlaps with the requirement that all sequences $k, 2k, \ldots, 12k$ fully populate each cycle before repeating, which a mere 12 subsets cannot accommodate without conflict due to their limited count.

Hence, a contradiction arises, and it is impossible to maintain such a partition under these conditions. Therefore, the answer is:
$$\boxed{\text{No}}$$