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

Let  $a_1\in N$$Let{a}_1,a_1+1,a_1+2,a_1+\mathrm{3........}{a}_1+99$

$\overline{X}=\frac{a_1+a_1+1+\mathrm{...............}+a_1+99}{100}=\frac{2\left(100a_1\right)+99\times 100}{100\times 2}=a_{11}+\frac{99}{2}$

$MD=\sum _{i=1}^{100}\frac{|x_i-\overline{X}|}{100}=2\frac{(\frac{99}{2}+\frac{97}{2}+\mathrm{.......}+\frac{1}{2})}{100}=25\forall a_1\in N$

S = contain natural number