Floor Function

First $x$ must be less than $5$ , since otherwise $x^{\lfloor x \rfloor}$ would be $3125$ which is greater than $1000$ .

Because $\lfloor x \rfloor$ must be an integer, we have to do the following:

For $\lfloor x\rfloor =0$ , $N=1$ , as long as $x\ne 0$ . This gives us $1$ value of $N$ .

For $\lfloor x \rfloor = 1$ , $N$ can be anything between $1^1$ and $2^1$ excluding $2^1$ .

Therefore, $N=1$ . However, we got $N=1$ in case $1$ so it got counted twice.

For $\lfloor x\rfloor =2$ , $N$ can be anything between $2^2$ to $3^2$ excluding $3^2$ . This gives us $3^2-2^2= 5N's$

For $\lfloor x \rfloor=3$ , $N$ can be anything between $3^3$ to $4^3$ excluding $4^3$ . This gives us $4^3-3^3=37 N's$ .

For $\lfloor x \rfloor=4$ , $N$ can be anything between $4^4$ to $5^4$ excluding $5^4$ . This gives us

$5^4-4^4=369 N's$ . We stop here since $x<5$ . Thus the possible answers for $N= 1+5+37+369=\boxed{412}$ .