【Question 4】
A cloth bag contains 10 small balls each of red, yellow, and green colors, all of the same size. The red balls are marked with the number "4", the yellow balls are marked with the number "5", and the green balls are marked with the number "6". Xiao Ming draws 8 balls from the bag, and their sum of numbers is 39. How many red balls could there be at most?
【Analysis and Solution】
If all 8 balls are red, the sum of the numbers is $4 \times 8=32$; replacing a red ball with a yellow ball increases the sum by $5-4=1$; replacing a red ball with a green ball increases the sum by $6-4=2$;
$$
(39-32) \div 2=3 \cdots \cdots 1 ;
$$

Therefore, 3 red balls should be replaced with 3 green balls, and 1 red ball should be replaced with 1 yellow ball; that is, 4 red balls, 1 yellow ball, and 3 green balls have a sum of 39; on the other hand, if there are 5 red balls, the sum of the numbers does not exceed $4 \times 5+6 \times 3=38$; hence, the maximum number of red balls possible is 4.