Dice with lines between faces I have a weird problem that I have no idea of how to even approach, much less solve. Here goes:
You have a regular polyhedron with n faces, with a number i from 1 to n. (I.e. a normal die.) Each face has curves on its surface that cross over onto another adjacent face. The number of such crossings is equal to i. A curve can terminate on a face or connect two edges of the face. For example:

Is this possible for any given value of n? If not, what values, if any, is it possible for?
 A: Let's ignore for a moment how the faces of the polyhedron are connected (which ones are neighbors) and exactly how the numbers are placed. (Also, there are not regular polyhedra for all $n$; I assume we can use other polyhedra too for part of my answer.)
Let's make a graph with $n$ nodes (each node represents a face), and we draw an edge between vertices if your curve goes from one face to the other. (Note, it does not matter whether the curve continues on to another face or terminates. For our purposes we will just assume all curves terminate.) We label the nodes $1...n$, and your problem requires us to connect node $i$ with $i$ other nodes. 
This is impossible when $n \equiv 1, 2 \pmod{4}.$ To see why, note there must be $1 + 2 + 3 + \cdots + n = n(n+1)/2$ edge endpoints, and so there must be $n(n+1)/4$ edges. But this is only an integer when  $n \equiv 3, 0\pmod{4}$, so the other cases are impossible.
Note, in particular, this means for a normal dice with 6 faces, it is impossible, no matter how we place the 6 numbers. 
Let's look now at the cases when it is possible, and now we see that how numbers are placed on the faces matters. For example, if we have a polehedron with 7 faces (a cube with one corner truncated, for example), and we put a 7 so that it neighbors 1, 2 and 3, it is impossible (we will be one edge short for 7).
Is there always some arrangement that can work?
If we do not worry about whether our face graph can actually be realized by an actual polyhedron, then we can always find a solution:
Pair odd nodes (there must be an even number, since $n \equiv 3, 0\pmod{4}$), and draw an edge between each pair. We now have left $0, 2, 2, 4, 4, ..., n - 1, n - 1$ (if $n \equiv 3 \pmod 4)$, or $0, 2, 2, 4, 4, , n-1, n-1, n$ (if $n \equiv 0 \pmod 4$).
In the first case, we simply draw edges between equal pairs. 
In the second, we match the last node (which as $n$ edges left) with the two nodes with $n/2$ edges left, and draw edges between equal pairs for the rest.
I demonstrate with $n = 7$ and $n = 8$. (I update the number to show how many edges still needs to be drawn in each step.)


This scheme works for the tetrahedron (since each face is connected to all the others). 
It can also work for the octahedron with suitable placement (in my picture, the green faces are the four at the top, and the red faces the four at the bottom):

I did not try the other 2 polyhedra; my guess it will work for both (with suitable placement of the numbers) since the network of the general solution is fairly "loose". You can try it. 
