1. Determine the smallest positive integer that can be inserted between the digits 20 and 16 so that the resulting number $20 \ldots 16$ is a multiple of 2016.
1. The number 2016 is a multiple of 9, so the resulting number must have a digit sum divisible by 9. This will occur precisely when the inserted number also has a digit sum divisible by 9, i.e., it must be a multiple of nine. We will try the positive multiples of the number 9 starting from the smallest: the numbers 9, 18, 27 do not work (instead of directly dividing by 2016, it is sufficient to verify that the numbers 20916, 201816, and 202716 are not — based on their last four digits — divisible by 16, whereas the number 2016 is), but $203616=2016 \cdot 101$. The smallest number sought is therefore 36.

Note. Since 2016 is a multiple of sixteen, the search for the smallest suitable number can also be based on the following obvious observation: a number with the last two digits 16 (i.e., a number of the form $100k+16$) is divisible by sixteen if and only if its penultimate two digits (i.e., the last two digits of the corresponding $k$) are divisible by four.

For a complete solution, award 6 points (even if it does not include considerations about divisibility by 9 or 16: it is possible, for example, to sequentially insert the numbers $1, 2, \ldots, 36$). For an incomplete solution: 3 points for justifying that the inserted number must be divisible by nine, respectively, four; 1 point for verifying that the number 36 meets the condition from the problem.