Arrangement of points in a circle From the 2015 Moscow Mathematical Olympiad:

The numbers $1$ to $1000$ are arranged on a circle such that each number divides the sum of its two neighbors. Suppose that the number $k$ has two odd neighbors. What's the parity of $k$ ?

I haven't been able to answer the question just yet. 
Here are a few things I noticed. The constraint "$x$ divides its two neighbors $a$ and $b$" gets tighter when $x$ grows. For example, if $x=1000$, we must have $1000=a+b$. If $x=999$, we have either $999=a+b$ or $1998=a+b$ (in that case $(a,b)\in \{(1000,998),(998,1000)\}$).
Nevertheless, I can't derive anything useful from $k| a+b$ with $a,b$ being odd numbers.
 A: diagram
the numbers from 1 to 1000 have equal number of even(E) and odd(O) numbers.
Let us assume an arrangement where there is at least one group of an odd number with two odd neighbors.This is arrangement A.
So we have 3 odds one after another. Let all evens in A have two odd neighbors. But then if we group two numbers in the anticlockwise direction we end up with an odd number for every even number, along with some odds extra.Thus it is not possible. 
Again let in A, we have some evens with one odd & one even neighbor instead of all evens with odd neighbors. But since even doesn't divide even+odd, so this arrangement too is futile.
Thus, for our given arrangement A, we should have an even number with 2 even neighbors, like: ......,E1,E2,E3,..... In such case,E3 must be followed by E4,
for otherwise we have ...,E1,E2,E3,O,.... but E3 can't divide E2+O. Again, for E4 we must have a consequent E5, and for E5 we must have E6 ;thus this makes all numbers on the circle even,which is impossible.
*Note that we have exhausted all variations of arrangement for A, each time getting a contradiction. Thus arrangement A is impossible.
Therefore, a number with two odd neighbors must be even.*
A: Just write out the numbers is their order from $1$ to $1000$ on the circle and the condition is satisfied. The answer to the question is now even. You will have $1000$ between $1$ and $999$ and every other number $n$ will have $n+1$ and $n-1$ whose sum is $2n$.
