Solution: We will prove that it is necessary to draw at least 66 balls. Let the white balls be no more than the balls of each of the other colors. Then the white balls are no more than 8 (otherwise, the balls of each color are no less than 9, and we can take 8 white, 9 blue, and 9 red, a total of 26 balls, among which there are no 10 balls of one color, a contradiction).

If the white balls are exactly 8, then there are exactly 8 balls of another color (otherwise, as above, we take $8+9+9=26$ balls and get a contradiction). Then it is sufficient to take $8+8+30=46$ balls: among them, there will definitely be 30 balls of one color.

If the white balls are no more than 7, then by taking any $7+29+30=66$ balls, among them there will be at least 59 red and blue balls, among which there will be at least 30 of one color.

We will show that 65 balls are not sufficient. Let the white balls be 7, the blue balls be 29, and the red balls be 64. The condition of the problem is satisfied: among any 26 balls, there are no fewer than $26-7=19$ blue and red balls, so at least 10 are of one color. On the other hand, by taking 7 white, 29 blue, and 29 red balls, we will take a total of 65 balls, among which there are no 30 of one color.
Problem 2. (3 points) In a box, there are white, blue, and red balls, totaling 100. If 26 balls are drawn from the box with closed eyes, there will definitely be 10 balls of the same color among them. How many balls must be drawn with closed eyes from the box to ensure that there are 30 balls of the same color among them?