At least how many groups do we need to divide the first 100 positive integers into so that no group contains two numbers where one is a multiple of the other?
Among two powers of 2, the smaller one is a divisor of the larger one, which means that out of the seven powers of 2 less than 100, no two can be placed in the same group. This implies that at least 7 groups are needed.

However, this is sufficient if the numbers $n$ that fall into the group of $2^{k}$ are such that $2^{k} \leqq n<2^{k+1}(0 \leqq k \leqq 6)$. The ratio of any two such numbers—when the larger is divided by the smaller—lies between 1 and 2, and thus cannot be an integer.

Remarks. 1. The above is not the only possible partition. If the numbers that have exactly $i$ prime factors in their prime factorization are placed in the $i$-th group, and 1 is placed in a separate group, we will get exactly 7 suitable groups. The smallest product of 7 factors is $2^{7}$, which is greater than 100, and a number has fewer prime factors in its proper divisors than in the number itself, so indeed no divisibility can hold among the numbers in the same group.

2. By a similar line of reasoning, it can be shown that the first $N$ positive integers can be placed into $1+\left[\log _{2} N\right]$ groups in the required manner, but not into fewer groups.