Let $$S$$ be the set of all values of $$a_1$$ for which the mean deviation about the mean of $$100$$ consecutive positive integers $$a_1, a_2, a_3, \ldots, a_{100}$$ is $$25$$. Then $$S$$ is

We need to find the set $$S$$ of all values of $$a_1$$ for which the mean deviation about the mean of $$100$$ consecutive positive integers $$a_1, a_2, \ldots, a_{100}$$ equals $$25$$.

The integers are $$a_1, a_1+1, a_1+2, \ldots, a_1+99$$, and their mean is $$a_1 + \frac{0 + 1 + 2 + \cdots + 99}{100} = a_1 + \frac{99 \times 100/2}{100} = a_1 + 49.5$$.

The mean deviation is $$\text{MD} = \frac{1}{100} \sum_{i=0}^{99} \bigl|a_1 + i - (a_1 + 49.5)\bigr| = \frac{1}{100} \sum_{i=0}^{99} |i - 49.5|$$.

The deviations are $$49.5, 48.5, 47.5, \ldots, 0.5, 0.5, 1.5, \ldots, 49.5$$, so $$\text{MD} = \frac{2}{100} (0.5 + 1.5 + 2.5 + \cdots + 49.5)$$.

This is an arithmetic series with $$50$$ terms, first term $$0.5$$, last term $$49.5$$, and its sum is $$\frac{50}{2}(0.5 + 49.5) = 25 \times 50 = 1250$$.

Therefore, $$\text{MD} = \frac{2 \times 1250}{100} = \frac{2500}{100} = 25$$.

The mean deviation is $$25$$ regardless of the value of $$a_1$$, so the set $$S$$ of all possible values of $$a_1$$ is the set of all natural numbers $$\mathbb{N}$$.

The correct answer is Option C: $$\mathbb{N}$$.