jmc

First, since $\lfloor x+y\rfloor ,$ $\lfloor x\rfloor ,$ $\lfloor y\rfloor$ are all integers, $$\lfloor x+y\rfloor -\lfloor x\rfloor -\lfloor y\rfloor$$must also be an integer.

We can write $$x=\lfloor x\rfloor +\{x\},$$where $\{x\}$ represents the fractional part of $x.$ Similarly, we can also write $y=\lfloor y\rfloor +\{y\}$ and $x+y=\lfloor x+y\rfloor +\{x+y\},$ so $$\begin{array}{rl}\lfloor x+y\rfloor -\lfloor x\rfloor -\lfloor y\rfloor & =(x+y-\{x+y\})-(x-\{x\})-(y-\{y\})\\ & =\{x\}+\{y\}-\{x+y\}.\end{array}$$Note that $0\le \{x\},$ $\{y\},$ $\{x+y\}<1,$ so $$\{x\}+\{y\}-\{x+y\}>0+0-1=-1$$and $$\{x\}+\{y\}-\{x+y\}<1+1-0=2.$$Since $\lfloor x+y\rfloor -\lfloor x\rfloor -\lfloor y\rfloor =\{x\}+\{y\}-\{x+y\}$ is an integer, the only possible values are 0 and 1.

For $x=y=0,$ $$\lfloor x+y\rfloor -\lfloor x\rfloor -\lfloor y\rfloor =0-0-0=0,$$and for $x=y=\frac{1}{2},$ $$\lfloor x+y\rfloor -\lfloor x\rfloor -\lfloor y\rfloor =1-0-0=1.$$Therefore, the possible values of $\lfloor x+y\rfloor -\lfloor x\rfloor -\lfloor y\rfloor$ are $\boxed{0,1}.$