your guess may be correct...too lazy to check though

all variables are integers mod 2n. In this case, n is 1004.
we call a sequence {x_1,x_2,..,x_n} good if {x_i +k} is disjoint from {x_i} for some k.

a good sequence corresponds to a possible arrangement, and can be treated as a binary sequence of length 2n by:
x_i = 0 if x_i not in X and 1 otherwise.
let the number of such binary sequence be f(2n) for a given n.

conjecture 1A(haha2000's 想法): k is divisible by n. consider a polynomials in x such that x^2n = 1. then sum of x^(x_i) + x^(x_i+k) = 1+x+x^2+...+x^2n = 0 implies that (sum of x^(x_i))(x^k+1) = 0. either x^2k = 1 or sum of x^(x_i) = 0. the latter case is too unlikely to be true... so k divides n.

conjecture 2: by 1, we seek to construct a good sequence X from a given k. if x_i is in X then x_i + k and x_i-k are not in X. so every element not in X is exactly covered by two elements in X.( i skipped some arguments here). hence, if x_i is in X then all x_i + 2mk are in X too, and all x_i+(2m+1)k are not in X. in other words, the sequence is uniquely determined by any n/k (or 2n/k? but you get the idea right)consecutive block.

conjecture 3: by 2, we see that f(2n) = sum of non periodical binary sequences of lengh d, where d divides n( or 2n...). but the number of non periodical binary sequence of length n, say a(n), is given by
a(n) = 2^n - sum of a(d) where d divides n. so we can actually compute f(2n) thru such relation.

i guess ettubrute's answer is along the right line, if conjecture 1 holds.