Digits product

The prime factorization of 2016 is $2016 = 2^5 \times 3^2 \times 7.$ Next, we will group the prime factors so that the product of each group is less than 10 i.e. they form digits. Since we are interested in the smallest positive integer with digital product 2016, we try to form as few groups as we can, so that we can minimize the number of digits in the integer.

After some experimentation, we find that the absolute smallest number of groups we can make is 4. The possibilities are

$2016 = (2 \times 2 \times 2) \times (2 \times 2) \times (3 \times 3) \times 7 = 8 \times 4 \times 9 \times 7 \\ 2016 = (2 \times 2 \times 2) \times (2 \times 3) \times (3 \times 3) \times 7 = 8 \times 6 \times 6 \times 7.$

A number is a multiple of 3 if and only if its digits sum to a multiple of 3. Since $8 + 4 + 9 + 7 = 26,$ a number composed of the digits 8, 4, 9, 7 would not be a multiple of 3. However, $8 + 6 + 6 + 7 = 27,$ which is a multiple of 3. Thus, a number composed of the digits 8, 6, 6, 7 would be a multiple of 3.

Finally, we list these digits in ascending order to produce the smallest positive integer whose digital product is 2016: $\boxed{6678}.$