Find all conditions for $x$ that the equation $1\pm 2 \pm 3 \pm 4 \pm \dots \pm n=x$ has a solution.
An attempt to "prove without words":
Ok, I'll give some words...
We start with the only sum $0$ : first row, rowindex $k=0$, set-of-results $R_0=\{0\}$ ).
After that we initialize our set of summands by $1$ and by subtraction (red arrow) or addition (green arrow) we arrive at the set of results $R_1 = \{-1,+1\}$ . Note the distance of all elements is $2$.
After that we add $2$ to our summands and by subtraction or addition to any element of the previous set of results we get $R_2=\{-3,-1,1,3\}$
After that we add $3$ to our summands and by subtraction or addition to any element of the previous set of results we get $R_3=\{-6,-4,-2,0,2,4,6\}$
...
We already see, how this generalizes:
The new summand $s_k$ subtracted from the smallest element in the previous result gives the new smallest element which is $- \binom{1+s_k}{2}$ (leftmost red arrow) and the analoguous to the old largest element gives $+ \binom{1+s_k}{2}$. Because the previous set of results was dense in steps of 2, and $s_k$ is smaller than the largest element of the previous result we get no new holes, and thus the result-set is again dense in steps of $2$, reaching from the negative binomial value to the positive binomial value.
Counting the cardinalities of the result-sets show, that we always have $ \operatorname{card}(R_k)= \binom{1+k}{2}+1 $ where $\operatorname{card}(R_0)=1$
The result for $R_{1395}$ should then be $$\operatorname{card}(R_{1395}) = \binom{1+1395}{2}+1 = 973711 $$
[Update] See my similar answer for the case that the first element of the sum is not $\pm 1$ but only $1$ .
After you've the second time changed your title (what's this for?) the other picture is relevant instead of the first given one. The logic is the same; the grey arrows are that now-invalid ones for the set-of-preferred-results $R_k$ :
The result for this version of $R_{1395}$ should then be $$\operatorname{card}(R_{1395}) = \binom{1+1395}{2}-2 = 973709 $$
Let $N:=1+2+\ldots+1395$, which is even. You've already found the necessary conditions that $x$ be even and $-N\leq x\leq N$. These conditions are also sufficient; to show this it suffices to choose the signs in the sum such that it sums to $x$, for every even $x$ with $-N\leq x\leq N$.
For $x=N$ we simply choose the $+$-sign everywhere. For $x$ even with $-N\leq x<N$ the difference $N-x$ is even, and so $\frac{N-x}{2}$ is an integer between $0$ and $N$. This means we can write $\frac{N-x}{2}$ as a sum of distinct integers from $1$ to $1395$, though not necessarily in a unique way (this is very easy to prove by induction). Now choose the $-$-sign at all these integers in stead of the $+$-sign; this is the same as subtracting $\frac{N-x}{2}$ from $N$ twice, so this way the sum sums to $N-2\cdot\frac{N-x}{2}=x$.
EDIT: The suggested claim that is very easy to prove by induction (on $n$) is the following:
For every natural number $n$, every integer from $1$ to $1+2+\ldots+n$ can be written as a sum of distinct integers from $1$ to $n$.
The base case is clear, and the induction step (from $n$ to $n+1$) only requires the observation that the last $n+1$ integers can be written in the form $x+(n+1)$ where $x\leq1+2+\ldots+n$.
EDIT: Only on a third reading did I notice that the sign of the $1$ can not be chosen. To this question I don't have anything better or more clear to say than David K has already said in his answer, and in particular in his linked answer.
In general, if $n$ is a positive integer, the minimum value of the sum $\pm 1 \pm 2 \pm 3 \pm \cdots \pm n$ is $-\frac{n(n+1)}{2}$, the maximum is $\frac{n(n+1)}{2}$, and all integers of the same parity between those values are possible sums. That is, if $\frac{n(n+1)}{2}$ is even, all possible values are even, and if $\frac{n(n+1)}{2}$ is odd, all possible values are odd.
This is based on an answer I posted to a very similar question several months ago; the main question in that case, like the original version of this question, did not have $\pm$ in front of the $1,$ but the version with $\pm1$ is a relatively trivial extension of the problem for which I gave a solution at the end of that answer.
Since the sums with $\pm$ in front of the $1$ are much simpler to show than the sums without that $\pm,$ I'll recapitulate the argument for the sum with $\pm1.$
If any of the $\pm$ signs in $\pm 1 \pm 2 \pm 3 \pm \cdots \pm n$ is negative, we can make a greater number by changing that sign to a positive. But if all signs are positive then any change will make the sum less. So the maximum value has all signs positive, and is $$ 1 + 2 + 3 + \cdots n = \frac{n(n+1)}{2}. $$ A mirror-image argument shows that the minimum value is $-\frac{n(n+1)}{2},$ achieved when all the $\pm$ signs are negative.
Now suppose that somewhere in the sum there are two consecutive terms whose signs are positive and then negative, that is, the terms are $+j - (j+1)$ where $1 \leq j \leq n-1.$ Then we can change these two terms to $-j + (j+1)$ in order to increase the sum by exactly $2.$ The only situation in which we cannot do this is when all the negative terms occur before all the positive terms; that is, either all are positive (the maximum value), all are negative (the minimum value), or the sum has the form $$ -1 - 2 - 3 - \cdots - k + (k+1) + (k + 2) + \cdots + n. $$ In the "all positive" case, of course we cannot increase the sum; in the other two cases, the first term is $-1,$ and we can change this to $+1$ and thereby increase the sum by $2.$
So every sum except the maximum one can be increased by $2$; starting at $-\frac{n(n+1)}{2},$ that lets us reach every integer of the same parity from $-\frac{n(n+1)}{2}$ to $\frac{n(n+1)}{2},$ inclusive.
You cannot reach any number of opposite parity from $\frac{n(n+1)}{2},$ because every time you change the sign of the term $j$ you either add or subtract $2j$ from the sum.
Now just plug in $1395$ for $n.$