The sequence $$1,2,3,...n$$ is taken ,if we take all the different possiblities of $+$ or $-$in between all successive terms ,what will be the minimum between all those (as in the minimum of the absolute value) .
My work:
I found that we can define a vector valued function that : $$f:\mathbb R^n \mapsto \mathbb R$$ and if we define $\mathrm v $to be the vector:$$(\pm1,\pm1,\pm1...,\pm1)$$ and the vector $\mathrm w$ to be:$$(1,2,3,...,n)$$ the function f is simply the dot product:$$ \vert v \cdot w\vert$$.
And also i found out that the function is multivalued(using examples),and that being multivalued has to do with number of partitions of a number ,other than that I have no clue on how to analyze this...
Lemma. A sequence of $4k$ consecutive positive integers can have $\pm$ signs assigned to its terms so that their sum becomes $0$.
Let the numbers be $m+1,m+2,\cdots,m+4k$. Each pair of symmetric terms $m+j,m+4k+1-j$ adds up to $m+j+(m+4k+1-j) = 2m+4k+1$. There are $2k$ such pairs, so assigning $k$ of them the $+1$ sign, and the other $k$ pairs the $-1$ sign, their sum will be $0$.
Proof. For an arbitrary $n$ let $k = \lfloor \frac{n}{4}\rfloor\,$ and choose signs per previous lemma for the last $4k$ terms so that they add up to $0\,$.
This leaves the first $n \bmod 4$ terms to consider.
If $n \bmod 4 = 0$ then there are no terms left, and the minimum sum is $0$.
If $n \bmod 4 = 1$ then the minimum sum is $+1 = 1$.
If $n \bmod 4 = 2$ then the minimum sum is $-1+2=1$.
If $n \bmod 4 = 3$ then the minimum sum is $-1-2+3=0$.
To prove that $1$ is in fact the minimum sum in the cases $n \bmod 4 = 1,2$ note that each sign change of one term $k$ modifies the sum by $2k$ i.e. an even number, so it preserves the parity of the sum. But the sum $\sum_{k=1}^n k$ is odd when $n \bmod 4 = 1,2$ (as can be easily verified) so there exists no combination of signs to bring the sum down to $0$, therefore $1$ is the best that can be achieved.