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.

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.
