Sum $\pmod{1000}$ 
Let $$N= \sum_{k=1}^{1000}k(\lceil \log_{\sqrt{2}}k\rceil-\lfloor \log_{\sqrt{2}}k \rfloor).$$ Find $N \pmod{1000}$.

Let $\lceil x \rceil$ be represented by $(x)$ and $\lfloor x \rfloor$ be represented by $[x]$. 
Consider $0 < x < 1$ then:
$$(x) - [x] = 1 - 0 = 1$$ 
Consider $x=0$ then:
$$(x) - [x] = 0$$
Consider $x=1$ then:
$$(x) - [x] = 1-1 = 0$$
So we must find $k$ such that:
$$\log_{\sqrt{2}}k \in \mathbb{Z}$$
$$\log_{\sqrt{2}}k = \frac{2\log k}{\log 2} = y \in \mathbb{Z}$$
$$\log k = \log(\sqrt{2}^y)\implies k = \sqrt{2}^y$$
So for a few integer $y$ values,
$$k = \sqrt{2}, 2^1, \sqrt{2}^3, 2^2,...,  2^3$$...
$k$ is an integer for powers of $2$. 
$$\log_{\sqrt{2}} 2^x = x\log_{\sqrt{2}} 2 = x \frac{2\log_2 2}{\log_2 2} = 2x \in \mathbb{Z}$$
The sum is $1$ for $k=1$ but then $0$ for $k=2$. 
$2^9 = 512 < 1000$ but $2^{10} = 1024 > 1000$.
but how can I compute the sum?

You've correctly identified two statements:
$\lceil x \rceil - \lfloor x \rfloor = \begin{cases} 0 & \text{ if } x \in \mathbb{Z} \\ 1 & \text{ if } x \notin \mathbb{Z} \end{cases}$
For integers $k$, $\log_{\sqrt{2}} k \in \mathbb{Z} \iff k = 2^r$ for some integer $r$.
Combining these, we get 
$\lceil \log_{\sqrt{2}} k \rceil - \lfloor \log_{\sqrt{2}} k \rfloor = \begin{cases} 0 & \text{ if } k = 2^r, \text{ some }r \\ 1 & \text{ otherwise } \end{cases}$
So
$$\sum_{k=1}^{1000}k(\lceil \log_{\sqrt{2}}k\rceil-\lfloor \log_{\sqrt{2}}k \rfloor) = \sum_{1 \leq k \leq 1000\text{, } k \neq 2^r\text{ for any } r \in \mathbb{Z}} k$$
$$ = \sum_{k=1}^{1000} k - \sum_{r=0}^92^r$$
which I'm pretty certain you can work out.
