Answer: 483

Solution:

On the strip $m \le x \lt m + 1$ we have $\lfloor\sqrt{x}\rfloor = k = \lfloor\sqrt{m}\rfloor,$ so the region above it is a trapezoid under $y = kx$ with area $A(m+1) - A(m) = \frac{(2m+1)k}{2}:$ an integer when $k$ is even, a half-integer when $k$ is odd. Hence as $n$ grows by $1,$ the integrality of $A(n)$ is unchanged while $k$ is even and flips at every step while $k$ is odd.

Consider the block of $2k + 1$ values $k^2 \lt n \le (k+1)^2.$ Starting from $A(1) = 0,$ the statuses of $A(k^2)$ cycle with period $4:$ integer for $k \equiv 0, 1$ and non-integer for $k \equiv 2, 3 \pmod 4$ (an odd block flips the status an odd number of times, an even block preserves it). Counting integer values of $A(n)$ inside each block: for $k = 4j - 3$ the block alternates, beginning and ending with non-integers, giving $4j - 3;$ for $k = 4j - 2$ every value is a non-integer, giving $0;$ for $k = 4j - 1$ it alternates, beginning and ending with integers, giving $4j;$ for $k = 4j$ all $8j + 1$ values are integers.

For $j = 1, \ldots, 7,$ covering $2 \le n \le 29^2,$ the four blocks contribute $(4j - 3) + 0 + 4j$ $+ (8j + 1) = 16j - 2$ integers, totaling $\sum_{j=1}^{7}(16j - 2) = 434.$ Then the block $k = 29$ contributes $29$ integers for $841 \lt n \le 900,$ the block $k = 30$ contributes none, and for $k = 31$ the alternation over $961 \lt n \le 1000$ begins with an integer at $n = 962$ and gives $20$ more. The total is $434 + 29 + 20 = 483.$