16. Let's consider the most "unfavorable" case: among the taken pencils, there could be 14 yellow, 11 brown, 11 black, 17 red, 17 green, and 17 blue. Thus, if we take 87 pencils $14+11+$ $+11+17+17+17=87$, it might happen that among them there will not be 18 pencils of the same color. By taking one more pencil, we will have either 18 red, 18 green, or 18 blue. Therefore, we need to take 88 pencils. In other cases, among the 88 taken pencils, there will be more than 18 pencils of one color.
16. In the box, there are 120 colored pencils: 35 red, 23 green, 14 yellow, 26 blue, 11 brown, and 11 black. What is the smallest number of pencils that need to be taken from the box in the dark (without seeing the pencils) to ensure that there are at least 18 pencils of one color among them?