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.
 A: As others have stated,
we want a partition of
$\{1, 2, ..., n+1\}$
into two sets with equal sums.
The sum is
$\frac{(n+1)(n+2)}{2}$.
If $n=4k$,
the sum is
$(4k+1)(2k+1)$
which is odd
and therefore impossible.
If $n = 4k+1$,
the sum is
$(2k+1)(4k+3)$
which is also odd,
and therefore impossible.
If $n = 4k+2$,
the sum is
$(4k+3)(2k+2)$,
so it is not ruled out,
and each sum must be
$(4k+3)(k+1)$.
if $n = 4k+3$,
the sum is
$(2k+2)(4k+5)$
which is also not ruled out,
and each sum must be
$(k+1)(4k+5)$.
Here are my solutions
for the not impossible cases.
For the $n=4k+2$ case,
the sum must be
$(4k+3)(k+1)
=(4k+4-1)(k+1)
=4(k+1)^2-(k+1)
=(2k+2)^2-(k+1)
$.
The square there suggests,
to me,
the formula for
the sum of consecutive odd numbers
$1+3+...+(2m-1)=m^2$,
so $1+3+...+(4k+3) = (2k+2)^2$.
If $k+1$ is odd,
remove it from the sum
so it is
$(2k+2)^2-(k+1)$.
If $k+1$ is even,
both $1$ and $k$ are odd,
so remove them from the sum.
In either case, we have the desired partition.
For the $n=4k+3$ case,
the sum must be
$(4k+5)(k+1)
=(4k+4+1)(k+1)
=4(k+1)^2+(k+1)
=(2k+2)^2+(k+1)
$.
Again,
$1+3+...+(4k+3) = (2k+2)^2$.
If $k+1$ is even,
add it to the sum
so it is
$(2k+2)^2+(k+1)$.
If $k+1$ is odd,
$k+2$ is even,
so remove $1$
and add $k+2$ to the sum.
In either case, we have the desired partition.
I do not know how many partitions
can be found.
A: 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
$$(0-1-2+3)+(4-5-6+7)+(8-9-10+11)=0.$$
A: This is rather similar to other answers, but since I liked the (re)formulation of the proof in a comments to a recent question about the same problem, let me summarize what was said there.


*

*If we notice that $4=k+(k-1)-(k-2)-(k-3)$, this can be used to prove by induction that if $n$ has this property, so has $n+4$.

*Together with the observation that the sum $1+2+\dots+n+(n+1)$ has to be even, which eliminates two residue classes modulo $4$, this gives us the full characterization of possible $n$'s.

A: 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 
$(n-2)-(n-1)-n+(n+1)=0 \implies 
[\pm1\pm2...\pm(n-3)]+[\pm(n-2)\pm(n-1)\pm n\pm(n+1)]=0$ 
