Hello aeroflix
Here's the solution to number 1.
Suppose we take $\displaystyle n$ natural numbers starting from $\displaystyle m$. Then their sum is: $\displaystyle S = \tfrac{1}{2}n(2m+n-1)$$\displaystyle =1000$
$\displaystyle \Rightarrow n(2m+n-1) = 2000$
$\displaystyle \Rightarrow m = \frac12\left(\frac{2000}{n}-n+1\right)$
$\displaystyle \Rightarrow n$ is a factor of $\displaystyle 2000$ and $\displaystyle \frac12\left(\frac{2000}{n}-n+1\right)$ is a natural number.
You'll find that this is satisfied for: $\displaystyle n=1, m = 1000$
$\displaystyle n=5,m=198$
$\displaystyle n=16,m=55$
$\displaystyle n=25,m=28$
I'll leave you to write down the resulting sequences of numbers.
Grandad