jmc

Let $k$ be a positive integer. Then for $k^2\le x<(k+1{)}^2,$ $$x\lfloor \sqrt{x}\rfloor =kx.$$Thus, on this interval, the graph of $0\le y\le x\lfloor \sqrt{x}\rfloor$ is a trapezoid, with left height $k^3,$ right height $k(k+1{)}^2,$ and base $(k+1{)}^2-k^2=2k+1,$ so its area is $$\frac{k^3+k(k+1{)}^2}{2}\cdot (2k+1)=2k^4+3k^3+2k^2+\frac{k}{2}.$$Let $n$ be a positive integer such that $k^2+1\le n\le (k+1{)}^2.$ Then for $k^2\le x<n,$ the graph of $0\le y\le x\lfloor \sqrt{x}\rfloor$ is a trapezoid with left height $k^3,$ right height $kn,$ and base $n-k^2,$ so its area is $$\frac{k^3+kn}{2}\cdot (n-k^2)=\frac{k(k^2+n)(n-k^2)}{2}=\frac{k(n^2-k^4)}{2}.$$We want to compute the area of the graph for $1\le x\le n$; in particular, we want this area to be an integer. We know that the area for $k^2\le x\le (k+1{)}^2$ is $$2k^4+3k^3+2k^2+\frac{k}{2}.$$Since $2k^4+3k^3+2k^2$ is always an integer, for our purposes, we keep only the $\frac{k}{2}$ term. This gives us $$\begin{array}{rl}\sum _{i=1}^{k-1}\frac{i}{2}+\frac{k(n^2-k^4)}{2}& =\frac{1}{2}\cdot \frac{(k-1)k}{2}+\frac{k(n^2-k^4)}{2}\\ & =\frac{k(k-1)}{4}+\frac{k(n^2-k^4)}{2}\\ & =\frac{k[2k(n^2-k^4)+k-1]}{4}.\end{array}$$Thus, we want $k[2k(n^2-k^4)+k-1]$ to be divisible by 4. We compute $k[2k(n^2-k^4)+k-1]$ modulo 4 for $0\le k\le 3$ and $0\le n\le 3,$ and obtain the following results:

$$\begin{array}{ccccc}k\backslash n& 0& 1& 2& 3\\ 0& 0& 0& 0& 0\\ 1& 2& 0& 2& 0\\ 2& 2& 2& 2& 2\\ 3& 0& 2& 0& 2\end{array}$$Case 1: $k=4m$ for some integer $m.$

All integers $n$ in the range $k^2+1\le n\le (k+1{)}^2$ work, for a total of $2k+1$ integers.

Case 2: $k=4m+1$ for some integer $m.$

Only odd integers $n$ in the range $k^2+1\le n\le (k+1{)}^2$ work. These are $k^2+2,$ $k^2+4,$ $\dots ,$ $(k+1{)}^2-1,$ for a total of $k$ integers.

Case 3: $k=4m+2$ for some integer $m.$

No integers $n$ in the range $k^2+1\le n\le (k+1{)}^2$ work.

Case 4: $k=4m+3$ for some integer $m.$

Only even integers $n$ in the range $k^2+1\le n\le (k+1{)}^2$ work. These are $k^2+1,$ $k^2+3,$ $\dots ,$ $(k+1{)}^2,$ for a total of $k+1$ integers.

Thus, the four cases $k=4m+1,$ $4m+2,$ $4m+3,$ and $4m+4$ contribute $$4m+1+4m+4+2(4m+4)+1=16m+14.$$integers.

Summing over $0\le m\le 6$ covers the cases $2\le n\le 841,$ and gives us $$\sum _{m=0}^6(16m+14)=434$$integers.

For $k=29,$ which covers the cases $529\le n\le 900,$ we have another 29 integers.

For $k=30,$ which covers the cases $901\le n\le 961,$ there are no integers.

For $k=31,$ only the even integers in the range $962\le n\le 1024$ work. We want the integers up to 1000, which are $$962,964,\dots ,1000,$$and there are 20 of them.

Thus, the total number of integers we seek is $434+29+20=\boxed{483}.$