<s>[INST] How many values does the expression $1 \pm 2 \pm 3 \pm \cdots \pm n$ take? 
How many different values does the expression $1 \pm 2 \pm 3 \pm \cdots \pm n$ take?

I was wondering about this problem and didn't think it was immediately obvious. The answer can't be $2^{n-1}$ since the combinations all might not be unique. I was thinking of looking at partitions of $\{1,2,\ldots,n, k\}$ where $1 \pm 2 \pm 3 \pm \cdots \pm n = k$ and thus the sum must be even. Therefore, $\dfrac{n(n+1)}{2}+k = \dfrac{n(n+1)+2k}{2}$ must be even which means $n(n+1) + 2k$ must be a multiple of $4$ so $n(n+1)+2k = 4m$. Thus if $n \equiv 1,2 \bmod 4$, then $k \equiv 1,3 \bmod 4$ and if $n \equiv 3,0 \bmod 4$ then $k \equiv 0,2 \bmod 4$.
 [/INST] Let us use generating functions
$$G_n(X)=X\prod_{k=2}^n(X^{-k}+X^k)=\dfrac{1}{X^p}\prod_{k=2}^n(1+X^{2k}) \ \ \ \text{with} \ \ \ p=\frac{n(n+1)}{2}-2$$
(very close to the proposal of @Jack D'Aurizio). In this way, we transfer the issue into counting how many combinations as positive or negative exponents of indeterminate $X$ are present in the expansion of $G_n(X)$.
Let us manipulate the first values of $n$ which is enough to get a good perception of the method. 
We begin by the case  $n=3$: 
$$G_3(X)=X(X^{-2} + X^2)(X^{-3} + X^3)=$$
$$X^{1-2-3} + X^{1+2-3} + X^{1-2+3} +X^{1+2+3}$$
$$=X^{-4} + X^0 + X^2 + X^6$$
Note the central symmetry (with respect to position $1$) of the exponents $-4,0,2,6$. This central symmetry will exist for all values of $n$.
Let us now consider the cases $n=4, 5, 6$:
$n=4$:
$$G_4(X)=X(X^{-2} + X^2)(X^{-3} + X^3)(X^{-4} + X^4) =$$
$$X^{-8} + X^{-4} + X^{-2} + X^0+ X^2 + X^4 + X^6 + X^{10}$$
Till now, there are no repetitions. Here they come for $n \geq 5$:
$n=5$:
$$G_5(X)=X(X^{-2} + X^2)(X^{-3} + X^3)(X^{-4} + X^4)(X^{-5} + X^5)$$
$$=X^{-13} + X^{-9} + X^{-7} + X^{-5} + 2X^{-3} + 
  X^{-1} + 2\,X + X^3 + 2\,X^5 + X^7 + X^9 + X^{11} + 
  X^{15}$$
This time, not all coefficients are one; the reason is clear: for example, monomial $^2X^5$ means that $5$ is obtainable in two ways, i.e., $1+2+3+4-5$ and $1-2-3+4+5$. This is an "added value" of the approach by generating functions.
A last case, $n=6$, in order to see the differences between the cases $n$ odd and $n$ even:
$$X(X^{-2} + X^2) \cdots (X^{-5} + X^5)(X^{-6} + X^6)=$$
$$=X^{-19} + X^{-15} + X^{-13} + X^{-11} + 2X^{-9} + 
  2X^{-7} + 2X^{-5} + 2X^{-3} + 
  3X^{-1} + 2\,X + 3\,X^3 + 2\,X^5 + 2\,X^7 + 2\,X^9 + 
  2\,X^{11} + X^{13} + X^{15} + X^{17} + X^{21}$$
I will not go beyond this point because there is a clear recurrence relationship, but developing it would look to the reader like paraphrasing the (very detailed) redaction of @David K .
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