To solve this problem, we need to find the difference between the maximum and minimum possible values of the sums of the three integers that Prem missed when calculating the sum of the first $n$ positive integers.

1. Sum Formula for the First $n$ Positive Integers: The sum of the first $n$ positive integers is given by: $S = \frac{n(n + 1)}{2}$

2. Prem's Calculation: Prem calculated the sum to be 975. Since he missed three integers, the correct sum $S$ should be 975 plus the sum of the three missed integers. Let the sum of the three missed integers be $x$. Then: $S = 975 + x$

3. Finding $n$: We need to find an integer $n$ for which the sum $S = \frac{n(n + 1)}{2}$ is close to 975, factoring in that three numbers were missed.

4. Approximating $n$: We start by estimating $n$ such that: $\frac{n(n + 1)}{2} \approx 975$ Solving this equation gives: $n(n + 1) \approx 1950$ Checking values around this estimate, we find that for $n = 45$: $\frac{45 \times 46}{2} = 1035$ and for $n = 44$: $\frac{44 \times 45}{2} = 990$ Since 975 is closer to 990, we conclude that $n = 44$ is likely the correct value.

5. Sum of Missed Integers: If $n = 44$, the correct sum $S$ should be 990. Therefore: $975 + x = 990$ Solving for $x$ gives: $x = 990 - 975 = 15$ Thus, the sum of the three missed integers is 15.

6. Maximum and Minimum Possible Values: To maximize the sum of three integers that add up to 15, we choose the highest integers possible within the first 44 positive integers. Conversely, to minimize the sum, we choose the smallest integers.

   • Maximum Sum: Use the integers 5, 6, and 4 (highest possible small values that sum to 15).
   • Minimum Sum: Use the integers 1, 2, and 12 (lowest possible values that sum to 15).

7. Calculating the Difference: Both combinations yield sums that add up to 15, so there is no variation in the sum of missed integers in this context. The difference is therefore:

   $15 - 15 = 0$