Let 
$N = \sum_{k = 1}^{1000} k ( \lceil \log_{\sqrt{2}} k \rceil  - \lfloor \log_{\sqrt{2}} k \rfloor )$
Find the [remainder](https://artofproblemsolving.com/wiki/index.php/Remainder) when $N$ is divided by 1000.  ($\lfloor{k}\rfloor$ is the [greatest integer](https://artofproblemsolving.com/wiki/index.php/Floor_function) less than or equal to $k$, and $\lceil{k}\rceil$ is the [least integer](https://artofproblemsolving.com/wiki/index.php/Ceiling_function) greater than or equal to $k$.)
The ceiling of a number minus the floor of a number is either equal to zero (if the number is an [integer](https://artofproblemsolving.com/wiki/index.php/Integer)); otherwise, it is equal to 1. Thus, we need to find when or not $\log_{\sqrt{2}} k$ is an integer.
The change of base formula shows that $\frac{\log k}{\log \sqrt{2}} = \frac{2 \log k}{\log 2}$. For the $\log 2$ term to cancel out, $k$ is a [power](https://artofproblemsolving.com/wiki/index.php/Exponent) of $2$. Thus, $N$ is equal to the sum of all the numbers from 1 to 1000, excluding all powers of 2 from $2^0 = 1$ to $2^9 = 512$. 
The formula for the sum of an [arithmetic sequence](https://artofproblemsolving.com/wiki/index.php/Arithmetic_sequence) and the sum of a [geometric sequence](https://artofproblemsolving.com/wiki/index.php/Geometric_sequence) yields that our answer is $\left[\frac{(1000 + 1)(1000)}{2} - (1 + 2 + 2^2 + \ldots + 2^9)\right] \mod{1000}$.
Simplifying, we get
$\left[1000\left(\frac{1000+1}{2}\right) -1023\right] \mod{1000} \equiv [500-23] \mod{1000} \equiv 477 \mod{1000}.$ The answer is $\boxed{477}$