If $\lfloor x \rfloor = k,$ then as $x$ runs over $[k, k+1)$ the value $x^k$ covers exactly the integers from $k^k$ to $(k+1)^k - 1$

Sum the counts $(k+1)^k - k^k$ for $k = 1,$ $2,$ $3,$ and $4;$ for $k = 5$ the values already exceed $1000$

Suppose $\lfloor x \rfloor = k$ for a positive integer $k.$ As $x$ runs over $[k, k+1),$ the value $x^k$ increases continuously from $k^k$ toward $(k+1)^k,$ so the attainable integers $N$ are exactly those with $k^k \le N \le (k+1)^k - 1:$ there are $(k+1)^k - k^k$ of them, and these ranges are disjoint for different $k.$ If $0 \lt x \lt 1,$ then $x^{\lfloor x \rfloor} = x^0 = 1,$ which is already attained. If $x \lt 0,$ then $\lfloor x \rfloor$ is negative; an odd exponent gives a negative value, while an even exponent gives a value strictly between $0$ and $1.$ Thus no $x \lt 1$ produces a new positive integer.

For $k = 1,$ $2,$ $3,$ and $4,$ the counts are $2 - 1 = 1,$ $9 - 4 = 5,$ $64 - 27 = 37,$ and $625 - 256 = 369,$ and every such $N$ is at most $624 \lt 1000.$ For $k = 5$ the smallest value is $5^5 = 3125 \gt 1000.$

The total is $1 + 5 + 37 + 369 = 412.$