jmc

Take cases on the value of $\lfloor x\rfloor$:

If $\lfloor x\rfloor <0,$ then $x^{\lfloor x\rfloor }$ can never be an integer. If $\lfloor x\rfloor =0$ (and $x\ne 0$), then $x^{\lfloor x\rfloor }=x^0=1$ regardless of the value of $x.$ Thus $N=1$ ($1$ value). If $\lfloor x\rfloor =1,$ then $1\le x<2,$ and $x^{\lfloor x\rfloor }=x^1=x,$ so we still only have $N=1$. If $\lfloor x\rfloor =2,$ then $2\le x<3,$ and $x^{\lfloor x\rfloor }=x^2,$ so we get $N=4,5,\dots ,8$ ($5$ values). If $\lfloor x\rfloor =3,$ then $3\le x<4,$ and $x^{\lfloor x\rfloor }=x^3,$ so we get $N=27,28,\dots ,63$ ($37$ values). If $\lfloor x\rfloor =4,$ then $4\le x<5,$ and $x^{\lfloor x\rfloor }=x^4,$ so we get $N=256,257,\dots ,624$ ($369$ values). If $\lfloor x\rfloor \ge 5,$ then $x^{\lfloor x\rfloor }\ge 5^5=3125>1000,$ which is too large.

Therefore, the number of possible values for $N$ is $1+5+37+369=\boxed{412}.$