I was reading about amorphous sets recently and I thought the idea was quite intriguing, but couldn't come up with any natural example of one (perhaps because the existence of one implies the falsehood of Choice).
Setting my sights lower to include only computable sets, I came to suspect that a "computably amorphous" set of natural numbers may exist. So far I haven't been able to come up with an example, but I feel like there might be one.
A starting point is that if the set is written in ascending order as a sequence of natural numbers, every number in the sequence after the nth (or n^2th, or whatever) may be required to be a multiple of n, thereby guaranteeing that any partition of the set into residue classes modulo n contains only one infinite class, namely zero.
But I have no idea how to go further to having all computable partitions of the set contain only one infinite member. Does anyone else know whether this is possible and if so how to achieve it?
This was answered in the comments; I'm answering here to move this off the "unanswered" queue. I've made this CW so I don't get reputation for others' work.
No such set exists.
The key point here is that if $X$ (WLOG, infinite) is computable then $X$ can be computably listed in order as $X=\{x_0<x_1<x_2<x_3<...\}$. More precisely: if the characteristic function of $X$ ($n\mapsto 1$ iff $n\in X$, $n\mapsto 0$ otherwise) is computable, then the principle function of $X$ ($n\mapsto$ $n$th smallest element of $X$) is computable. This is a good exercise, and the converse holds too.
With this in hand, we can argue as follows. If $X$ is computable, let $\pi_X$ be its principle function. Then we can consider the set $$E_X:=\{\pi_X(2k):k\in\mathbb{N}\}.$$ We have that both $E_X$ and $X\setminus E_X$ are infinite, and $E_X$ is computable: to check if $n\in E_X$, just compute $\pi_X(0),\pi_X(2),\pi_X(4),...,\pi_X(2\cdot \lfloor {n\over 2}\rfloor)$ and check if $n$ is one of these values (note that if $n=\pi_X(2k)$ then $2k\le n$).