jmc

We will begin with the smallest possible positive values of $x$. For positive values of $x$, when $0<x<1$, the right side of our equation is equal to $\frac{1}{0}$, which is undefined. When $1\le x<2$ , the right side of our equation is equal to $1$, but $x-\lfloor x\rfloor$ cannot equal $1$.

When $2\le x<3$, the right side of our equation is equal to $\frac{1}{2}$, so we have $x-\lfloor x\rfloor =\frac{1}{2}$. This occurs when $x=2\frac{1}{2}$.

When $3\le x<4$, the right side of our equation is equal to $\frac{1}{3}$, so we have $x-\lfloor x\rfloor =\frac{1}{3}$. This occurs when $x=3\frac{1}{3}$.

When $4\le x<5$, the right side of our equation is equal to $\frac{1}{4}$, so we have $x-\lfloor x\rfloor =\frac{1}{4}$. This occurs when $x=4\frac{1}{4}$.

Then the sum of the three smallest positive solutions to $x$ is $2\frac{1}{2}+3\frac{1}{3}+4\frac{1}{4}=\boxed{10\frac{1}{12}}.$