TITLE: Find all conditions for $x$ that the equation $1\pm 2 \pm 3 \pm 4 \pm \dots \pm n=x$ has a solution.
QUESTION [13 upvotes]: 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.

REPLY [4 votes]: Let $K = \sum_{i=1}^n {k_i}i$ where $k_i \in \{1,-1\}$ be one of the expressible numbers and $M = \sum_{i=1}^n {m_i}i$ where $m_i \in \{1,-1\}$ be another.
$M - K = \sum_{i=1}^n (m_i - k_i) i = \sum_{i=1}^n [\{-2|0|2\}]i$ is an even number so all such numbers have the same parity. 
Cleary any $K$ is such $-\frac{n(n+1)}2=- \sum i \le K \le \frac{n(n+1)}n$.
Let $K <  \frac{n(n+1)}n$ so one of the ${k_i} = -1$.  Let $j$ be so that ${k_{m }} = 1; \forall m < j$ but ${k_j} = -1$.  
Let $\overline{K} = \sum {m_i}$ where $m_i = k_i$ for $i \ne j|j-i$; ${m_j} = 1$ (whereas ${k_j} = -1)$ and, if $j > 1$ then ${m_{j-1}} = -1$ whereas ${k_{j-1}} = 1$.  Then $\overline {K} = K + 2j - 2(j-1) = K + 2$.
So via induction, all (and only) $K;  -\frac{n(n+1)}2=- \sum i \le K \le \frac{n(n+1)}n; K $ of the same parity of $\frac{n(n+1)}n$ are possible.
So for $n = 1395$, all even numbers between $-\frac{1395*1396}2$ and $\frac{1395*1396}2$ are possible.