Divide a circle into $2n$ equal sections. We call a circle filled if it is filled with the numbers $0,1,2,\dots,n-1$. We call a filled circle good if it has the following properties:
$i$. Each number $0 \leq a \leq n-1$ is used exactly twice
$ii$. For any $a$ we have that there are exactly $a$ sections between the two sections that have the number $a$ in them.
Here is an example of a good filling for $n=5$ (View attachment)
Prove that there doesn’t exist a good filling for $n=1399$
Color the segments in black and white, alternating the colors.
Assume the contrary,
$i)$ call a number monochromatic if both segments containing the number are the same color.
$ii)$ call a number colorful if the segments containing said number have different colors.
It is evident that all odd numbers are monochromatic and even numbers are colorful.
$iii)$ call a monochromatic number black if its segments are black.
Suppose the number of black numbers are $M$.
On the other hand the numbers we have written on a filled circle of order $1399$ are $0$ to $1398$. Therefore we have written $700$ even numbers.
in this case by counting the number of all black segments we conclude that $1399=2M+700$
Which is obviously impossible for parity reasons
Label the segments from $1$ to $2n$ in clockwise direction.
Now assume that for some $n$ we can construct a good filled circle.
Note that the conditions implies that we have partitioned the set $\{1,\dots,2n\}$ into $n$ pairs with the following form
$$\{x_1,x_1+1\},\{x_2,x_2+2\},\dots,\{x_n,x_n+n\}\quad (\text{Values are considered}\ mod\ 2n)$$This comes from noting that if the first segment having $i$ in the clockwise direction is $x_i$ the next one is $x_i+i$ and the it's a partition of the set because each number appears exactly twice.
Now note that since $2n$ is even, the sum of both sides from this partition must be of the same parity, $i.e.$
$$1+2+\dots+2n\equiv (x_1+x_1+1)+(x_2+x_2+2)+\dots+(x_n+x_n+n)\quad (mod\ 2)$$This is equivalent to
that for $n\equiv 3\quad(mod\ 4)$, the left hand side of the above congruency is odd while the right hand side is even, and since $1399\equiv3\quad(mod\ 4)$, we conclude the result.
$$Q.E.D$$
Another solution by a friend of mine (Parsa Naderian) ,this is the exact thing he sent me:
FTSOC assume that we have a good filling for $1399$. Now start by adding the lines between the two $i$ Sections and numbering them and define by $x_i$ the number of the intersections of the $i_{th}$ line and the new lines drawn.
We know that there are $i+1$ lines in between the two sections with number $i$ in them so the line between the two $i$s will intersect $i+1$ lines
$$\implies \quad x_1+x_2+\dots+x_{1399}=1+2+3+\dots+1399= \text{even}$$So we find that at least one of the $x_i$s are even . We know that a new line intersects a line drawn before When the two numbers written next to the line are equal so there are even numbers that are written in the two sides of the line numbered $1$ .So there are $1399-x_i$ numbers written on one side of the line and they should be paired in equal numbers but it is impossible since $1399-x_i$ is odd.
so here is my solution , for the sake of contradiction assume the answer is yes , lets make a graph with $2798$ vertices, now between ani two vertice like $i$ draw $i$ clocl wise edges from one of them to all the other $i$ vertices between them , the number of edges is equal to $0+1+2+... +1398=1399(699)$ which is odd now take ani two numbers like $j,i$ and you can see easily that the edges between two $i$ and two $j$ is eather 2 or 0 which is even and by taking modulo 2 we get a contradiction
