Call $S_n$ the set of all values that ca be obtained with the sum
$$
1\pm2\pm3...\pm n
$$
with the convention that $S_1=\{1\}$.
You obtain $S_2$ by taking the sole element of $S_1$ and either adding or subtracting $2$:
$$
S_2=\{-1,3\}
$$
Similarly, you obtain $S_3$ by taking the elements of $S_2$ and either adding or subtracting $3$:
$$
S_3=\{-4, 0, 2, 6\}
$$
You do the same for $S_4$, taking the elements of $S_3$ and either adding or subtracting $4$:
$$
S_4=\{-8, -4, -2, 0, 2, 4, 6, 10\}.
$$
In general, you obtain $S_n$ but adding or subtracting $n$ to the elements of $S_{n-1}$. Delete duplicates if there are some. Keep going for a while:
\begin{align}
S_5&=\{-13, -9, -7, -5, -3, -1, 1, 3, 5, 7, 9, 11, 15\} \\
S_6&=\{-19, -15, -13, -11, -9, -7, -5, -3, -1, 1, 3, 5, 7, 9, 11, 13, 15, 17, 21\}\\
...
\end{align}
You can see the pattern: for $n\geq3$,
$$
S_n=\{min, min+4,min+6,...,-1,1,...,max-6,max-4,max\}.
$$
That is, all odd numbers from $min$ to $max$ except $min+2$ and $max-2$.
As others have pointed out,
\begin{align}
max&=1+2+...+n=\frac{n(n+1)}{2}\\
min&=1-2-3...-n=2-(1+2+...+n)=2-max.
\end{align}
When $n=1395$ you get
\begin{align}
max&=\frac{1395\times(1395+1)}{2}=973710\\
min&=2-max=-973708.
\end{align}
Therefore the conditions you are looking for in the case $n=1395$ are
- $x$ is odd
- $-973708\leq x\leq 973710$
- $x$ is neither $-973706$ nor $973708$.
Note. If you allow the first term to be $\pm1$ then $S_1=\{-1,1\}$. If you work ou the example you find the same values for $x$ except that $min+2$ and $max-2$ are not excluded.