Absolutely minimum

By the triangle inequality, we have $|x-1|+|100-x|\geq 99, |x-2|+|99-x|\geq 97,$ $...,(|x-50|+|51-x|\geq 1$ , so that the given sum is $\geq 1+3+...+97+99=50^2=\boxed{2500}$ . Equality is attained for $50\leq x\leq 51$ .

Let f(x) denote the given function. The problem can definitely be done by breaking into intervals but could be done more easily if we take the interpretation of the absolute value as the $distance$ . The given function can be interpretated as distances of x from 1,2,3...,100. On Real Number line, going towards left of 1 or right of 100 only increases distance of x from 1,2...,100, therefore f(x) will have minimum value somewhere between 1 and 100. Now as we move away from 1 towards 100, distance from 1,2,3.. increases while from 100,99,98... decreases. From symmetry one can easily conclude that f(x) will be minimum somewhere between 50 and 51{For 50≤x≤51 f(x) is a constant!!}. Hence f(x) minimum is 2500.

{ $2\displaystyle \sum_{r=1}^{100}r-\displaystyle \sum_{r=1}^{50}r=2500$ }

For x=1 and x=100, you get the same answer. For x=2 and x=99, you also get the same answer. Using the symmetry, x is 50.5 for it to be minimal. Hence, (49.5 +0.5)*50=2500.

Absolute deviation taken from mean of 1,2,3...….1000. I.e.50.50 is minimum