The problem states that we have $n$ blue, yellow, and red balls each, making a total of $3n$ balls in the bag. After drawing the 31st ball, all remaining balls are red. We are tasked with finding the difference between the maximum and minimum possible values of $n$.

Key observations:

1. There are $n$ red balls in total.
2. After drawing 31 balls, all remaining balls are red, meaning that all blue and yellow balls (combined total of $2n$ balls) must have been drawn within the first 31 draws.
3. Thus, out of the first 31 draws, all $2n$ blue and yellow balls were drawn, and the rest of the $31 - 2n$ balls must have been red.

Condition:

For the problem to hold true, we need $31 \geq 2n$, as no more than 31 balls are drawn.

This inequality simplifies to: $n \leq \frac{31}{2} = 15.5$ Since $n$ must be an integer, the maximum value of $n$ is 15.

Additionally, we know that $2n$ blue and yellow balls must be drawn within 31 balls. The minimum value of $n$ must be such that there are at least some red balls left after 31 draws. Thus, $n$ must be at least 1.

Conclusion:

• Maximum value of $n$ is 15.
• Minimum value of $n$ is 1.

Thus, the difference between the maximum and minimum values of $n$ is: $15 - 1 = 14$

The difference is $\boxed{14}$.

Would you like any further details or explanations?

Related questions:

1. What is the general formula for the number of balls remaining after a certain number of draws?
2. How can the number of each colored ball drawn be calculated in a similar problem with different conditions?
3. What if the condition required all remaining balls to be blue instead of red?
4. How does the total number of balls change if we have unequal numbers of blue, yellow, and red balls?
5. How would the solution differ if there were an additional color of ball?

Tip:

When dealing with probability or combinatorics problems, always write out the key constraints clearly before attempting the solution.