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.

Let $k=\sum_{i=1}^{1395}i=\frac 121395\cdot 1396$, which is the maximum sum you can attain.  Claim: you can achieve any even sum from $-k+2$ to $k$ except $k-2, -k+4$  We proceed by strong induction over numbers of the form $3 \bmod 4$.  The base cases are $n=3,$ where we should be able to achieve $6,2,-4$ which we can do with $1+2+3, 1-2+3, 1-2-3$ and $n=7$ where we can achieve $-26,-22,-20,-18,\ldots 24,28$.  Then we show if it is true up to $m$, it is true for $m+4$.  If our target is within the range we can obtain with $m$, we can put plus signs before $m+1, m+4$ and minus signs before $m+2, m+3$ and use the solution for the target with $m$.  If it is greater than $\frac 12m(m+1)$ it is less than or equal to $\frac 12(m+4)(m+5)=\frac 12m(m+1)+4m+10$  For $m \ge 11$ we can negate all the top four terms and reduce our target with numbers up to $m-4$ by $4m-10$.  This will not reduce the target below $0$ so we can use the solution for $m$ and the new target.  A similar argument works for negative targets with plus signs before the four largest numbers.
