For what $n$ can $\pm 1\pm 2\pm 3 ... \pm (n-1) \pm n = n+1$? More explicitly,
for what values of $n$
can the signs be chosen
in the equation
$$\pm 1\pm 2\pm 3 ... \pm (n-1) \pm n = n+1$$
so the equation is true?

For all $n$,
if the equation cannot be satisfied,
prove it;
if the equation can be satisfied,
exhibit a solution.

Note:
I have a solution,
and will present it in two days
if no one else submits a similar one.
Equivalently, we want to divide the set $\{1,2,3,\dots, n+1\}$ into two parts, such that the sums of the two parts are the same. The $n+1$ is somewhat unattractive, so let us call it $m$. 

A necessary condition is that the sum $1+2+3+\cdots+m=\frac{m(m+1)}{2}$ is even. We show that condition this is also sufficient. 

Our necessary condition is met precisely when $m$ is of the form $m=4k$ or of the form $m=4k-1$.  

Note that $(4j+1)+(4j+4)=(4j+2)+(4j+3)$. That takes care of all cases where $m$ is of the form $4k$. For any group $\{4j+1,4j+2,4j+3,4j+4\}$ can be divided into $2$ parts with equal sum.

For $m$ of the form $4k-1$, think pf our sum as starting at $0$, and use as groups $\{4j,4j+1,4j+2,4j+3\}$, dividing each group by using the fact that $4j+(4j+3)=(4j+1)+(4j+2)$. 

Remark: I feel guilty about these $j$ and $k$ and $m$ and $n$. It all comes down to two sample cases:
$$(1-2-3+4)+(5-6-7+8)+(9-10-11+12)=0,$$
and
$$(0-1-2+3)+(4-5-6+7)+(8-9-10+11)=0.$$
all n such that $n\equiv 3 \mod4$ can satisfy this.  and $\forall n$ that satisfy it $\exists k\in\mathbb{N}$ s.t. $n+4*k$ also satisfies it.  I say that because I am not 100% certain $n\equiv3 \mod 4$ is the entire set of solutions.

$\pm1\pm2...\pm n =n+1 $ gives us $\pm1\pm2...\pm(n+1)=0$ 
by induction if $n-3$ satisfies the second equation 
$(n-2)-(n-1)-n+(n+1)=0 \implies 
[\pm1\pm2...\pm(n-3)]+[\pm(n-2)\pm(n-1)\pm n\pm(n+1)]=0$