Find all conditions for $x$ that the equation $1\pm 2 \pm 3 \pm 4 \pm \dots \pm n=x$ has a solution. Find all conditions for $x$ so that the equation $1\pm 2 \pm 3 \pm 4 \pm \dots \pm 1395=x$ has a solution.
My attempt: $x$ cannot be odd because the left hand side is always even then we have $x=2k(k \in \mathbb{N})$ also It has a maximum and minimum 
$1-2-3-4-\dots-1395\le x \le 1+2+3+4+\dots +1395$
But I can't show if these conditons are enough or make some other conditions.

Given that the $1$ in the series is not $\pm1$, I can see one other condition, namely that $x$ cannot have the values $Max-2$ or $Min + 2$, where Max and Min are the maximum and minimum values of the series.
The Max occurs when all signs are positive. The only way to get a sum of $Max-2$ is by changing the sign of $1$ from positive to negative, thus decreasing the sum by $2$. Changing the sign of $2$ from positive to negative would decrease the sum by $4$. Changing the sign of $3$ from positive to negative would decrease the sum by $6$, etc.  It is thus not possible to achieve the number $Max - 2$.
The Min occurs when all signs are negative, except for the $1$ which must be positive. The smallest positive increase it is possible to make is by changing the sign of the $2$ from negative to positive. This increases the sum by $4$. It is thus not possible to achieve the number $Min + 2$. 
