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.

I am not sure this is the approach, but let me try. I am assuming $\pm1$ in the statement of the problem.
The basic idea is induction. For $n$, solve your problem. That is, $\pm1\pm2\ldots\pm n=x$. Then go to $n$.
Basic induction step: $n=3$. Then $\pm1\pm2\pm3=x$ has a solution for any even $x$ between $\pm6$. This is easy to check.
The induction step: given $n$ such that $n+1$ is divisible by $4$, go to $n+4$.
Let me illustrate this from $n=3$ to $n'=7$ as an example (hoping general step can be written). This involves adding $\pm4\pm5\pm6\pm7$. Using $4+7-5-6=0$, if there is a solution for $x$ with $n$, there is a solution for $x$ with $n'$. The $\pm4\pm5\pm6\pm7$ can moreover be used to add $\{2,4,8,10,12,14,22\}$ (to $6$ generated by $1+2+3$, subtracting is the same). If I show how to add $\{6,16,18,20\}$ the 'induction step' is complete. To add $6$ add $12$ with $\pm4\pm5\pm6\pm7$ and change $3$ to $-3$. To add $16$ add $22$ with $\pm4\pm5\pm6\pm7$ and change $3$ to $-3$. Likewise, $18$ and $20$ follow by adding $22$ and changing sign of $2$ and $1$ respectively.
