1. Write the 100 consecutive positive integers

   Since the numbers are consecutive, $a_1, a_2, a_3, \dots, a_{100} = a_1, a_1+1, a_1+2, \dots, a_1+99.$

2. Find their mean

   For an arithmetic progression, the mean is the average of the first and last terms: $\bar{x} = \frac{a_1 + (a_1+99)}{2} = a_1 + \frac{99}{2} = a_1+49.5.$

3. Compute the mean deviation about the mean

   Mean deviation about mean is $\text{MD} = \frac{1}{100}\sum_{k=0}^{99} \left| (a_1+k) - (a_1+49.5) \right|.$

   The term $a_1$ cancels: $\text{MD} = \frac{1}{100}\sum_{k=0}^{99} |k-49.5|.$

   So the mean deviation depends only on the number of terms and the common difference, not on $a_1$.

4. Evaluate the sum

   Distances from $49.5$ are: $49.5, 48.5, 47.5, \dots, 0.5, 0.5, \dots, 47.5, 48.5, 49.5.$

   Hence, $\sum_{k=0}^{99} |k-49.5| = 2(0.5+1.5+2.5+\cdots+49.5).$

   Now, $0.5+1.5+2.5+\cdots+49.5$ is an AP with 50 terms, first term $0.5$, last term $49.5$.

   Its sum is $\frac{50}{2}(0.5+49.5)=25\cdot 50=1250.$

   Therefore, $\sum_{k=0}^{99} |k-49.5| = 2\cdot 1250 = 2500.$

   So, $\text{MD} = \frac{2500}{100} = 25.$

5. Interpretation

   The mean deviation is always $25$, regardless of the value of $a_1$, as long as the 100 integers are positive.

   Since all numbers must be positive integers, we need $a_1 \in \mathbb{N}.$

   Thus, $S = \mathbb{N}.$

6. Check options

   • A: $\{9\}$ — incorrect
   • B: $\phi$ — incorrect
   • C: $\{99\}$ — incorrect
   • D: $\mathbb{N}$ — correct

Therefore, the correct option is D.