jmc

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 $$\begin{array}{rl}|x-1|+|x-1.5|+|x-2|& =(x-1)+(x-1.5)+(-(x-2))\\ & =2x-2.5-x+2\\ & =x-\mathrm{0.5.}\end{array}$$ 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 $$\begin{array}{rl}|x-1|+|x-1.5|+|x-2|& =(x-1)-(x-1.5)-(x-2)\\ & =x-1-x+1.5-x+2\\ & =-x+\mathrm{2.5.}\end{array}$$ 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 $$\begin{array}{rl}|x-1|+|x-1.5|+|x-2|& =-(x-1)-(x-1.5)-(x-2)\\ & =-3x+\mathrm{4.5.}\end{array}$$ 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 $y=|x-1|+|x-1.5|+|x-2|$.