No, that's not possible

Take the 500-th number as $k$ . This is the average of the 999 numbes.

Now,

$\sum_{i=0}^{499} (k-i)^2 = \sum_{i=1}^{499} (k+i)^2$

Thus,

$k^2 = \sum_{i=1}^{499} \left((k+i)^2-(k-i)^2\right)$

$k^2 = \sum_{i=1}^{499} (2k)(2i)$

$k = 4 \sum_{i=1}^{499} i = 4 \frac{499*500}{2} = \boxed{499000}$

$\text{Let the middle number be x, and the first 500, }\\ (x-499), (x-498), . . .(.x-1), x~~\\and~ the~ last~ 499, ~~~~(x+1),(x+2), . . . (x+499)\\ \text{Squaring adding and equating,}\\ (x-499)^2+ (x-498)^2+ . . . +(x-1)^2+ x^2\\=(x+1)^2+(x+2)+ . . . (x+499)^2\\Reduces~~ to,\\x^2=4x*(499+498+ . . . 2+1)=4x*\dfrac{499*500} 2= 499000x\\\implies~~the ~average=middle ~term ~=x=~~~~\large \color{#D61F06}{499000}$

Let $n$ be the least of the sequence of $999$ integers. Then the average of these numbers will be $n + 499.$ Now we are given that

$\displaystyle\sum_{k=n}^{n + 499} k^{2} = \sum_{k=n+500}^{n+998} k^{2}$

$\Longrightarrow (n + 499)^{2} = \displaystyle\sum_{k=n}^{n+498} ((k+500)^{2} - k^{2})$

$\Longrightarrow (n + 499)^{2} = \displaystyle\sum_{k=n}^{n+498} (1000k + 500^{2})$

$\Longrightarrow (n + 499)^{2} = \left(1000\displaystyle\sum_{k=n}^{n+498} k\right) + 499*500^{2}$

$\Longrightarrow (n + 499)^{2} = 500((n + 498)(n + 499) - (n - 1)n) + 499*500^{2} =$

$500(998n + 498*499) + 499*500^{2} =$

$500*998*(n + 249) + 499*500^{2} =$

$500*998*(n + 249 + 250) = 500*998*(n + 499)$

$\Longrightarrow n + 499 = 500*998 = \boxed{499000}.$