We tackle this with some careful casework. Case 1: $ x\ge 2 $. We then have $ |x-1| + |x-1.5| + |x-2| = (x-1) + (x-1.5) + (x-2) = 3x - 4.5 $. Since the smallest value of $ x $ in this case is $ 2 $, the smallest possible value of the sum in this case is $ 3(2) - 4.5 = 1.5 $. Case 2: $ 1.5\le x < 2 $. We then have \begin{align*} |x-1| + |x-1.5| + |x-2| & = (x-1) + (x-1.5) + (-(x-2)) \\ & = 2x - 2.5 -x+2 \\ & = x -0.5.\end{align*} Since the smallest possible value of $ x $ in this case is $ 1.5 $, the smallest possible value of the sum in this case is $ 1.5-0.5 = 1 $. Case 3: $ 1 \le x < 1.5 $. We then have \begin{align*} |x-1| + |x-1.5| + |x-2| & = (x-1) - (x-1.5) - (x-2) \\ & = x-1 -x + 1.5-x+2 \\ & = -x +2.5.\end{align*} Since $ x $ is less than 1.5, the sum in this case is greater than $ -1.5+2.5 = 1 $. Case 4: $ x < 1 $. We then have \begin{align*} |x-1| + |x-1.5| + |x-2| & = -(x-1) - (x-1.5) - (x-2) \\ & = -3x + 4.5.\end{align*} Since $ x $ is less than 1, the sum in this case is greater than $ -3(1) + 4.5 = 1.5 $. Reviewing these cases, we see that the least possible sum is $ \boxed{1} $. As an extra challenge, see if you can find a quick solution to this problem by thinking about the graph of $ y = | x-1| + |x-1.5| + |x-2| $.