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Step-by-Step Solution
Step 1: Express the 100 Consecutive Positive Integers
Let the first integer be $a_1 = a$. Then the 100 consecutive integers can be written as:
$a_1 = a,\quad a_2 = a+1,\quad a_3 = a+2,\quad \dots,\quad a_{100} = a+99.$
Step 2: Calculate the Mean of These 100 Integers
The mean $ \mu $ is given by:
$$
\mu \;=\;\frac{(a + (a+1) + (a+2) + \dots + (a+99))}{100}.
$$
We can factor out $a$ and sum the consecutive numbers:
$$
\mu \;=\;\frac{100\,a + (0 + 1 + 2 + \dots + 99)}{100}.
$$
The sum of the first 100 integers starting from 0 up to 99 is
$$
0 + 1 + 2 + \dots + 99 \;=\;\frac{99 \times 100}{2} = 4950.
$$
Hence,
$$
\mu \;=\; \frac{100a + 4950}{100} \;=\; a + 49.5.
$$
Step 3: Write the Mean Deviation About the Mean
The mean deviation about the mean (M.D.) of a set of values $x_1, x_2, \dots, x_n$ with mean $\mu$ is defined as:
$$
\text{M.D.} \;=\; \frac{1}{n} \sum_{i=1}^{n} |\,x_i - \mu \,|.
$$
For our sequence, $x_i = a + (i-1)$ for $i = 1, 2, \dots, 100,$ and $\mu = a + 49.5.$ Thus,
$$
\text{M.D.} \;=\; \frac{1}{100} \sum_{i=1}^{100} \bigl|\, (a + (i-1)) - (a + 49.5) \bigr|.
$$
Notice $a$ cancels out:
$$
\text{M.D.} \;=\; \frac{1}{100} \sum_{i=1}^{100} |\, (i-1) - 49.5 \,|.
$$
Step 4: Observe That the Mean Deviation Does Not Depend on $a$
Since $a$ cancels out, the expression depends only on the fixed set of distances from 0 through 99 to 49.5. As a result, this sum (and hence the mean deviation) is the same for any starting positive integer $a.$
Step 5: Conclusion About the Set $S$
The problem states that this mean deviation is 25. We see that the mean deviation for 100 consecutive integers will always be 25, regardless of the value of $a$ (provided $a$ is a positive integer). Therefore, the set $S$ of all possible $a$ for which the mean deviation is 25 is simply all positive integers. Symbolically:
$$
S = \{ a \in \mathbb{N} : a \ge 1 \}.
$$
This corresponds to the option given as “$N$”.
