Let $e=[1,1,\ldots,1]\in\mathbb{Z}^n$. I am looking for a way to find a vector $a\in\mathbb{Z}^n$ such that:


1. $\langle a,e\rangle=0$ and
2. for every nonnegative $v\in\mathbb{Z}^n$ such that $\langle e,v\rangle=n$ and $v\neq e$ we have $\langle a,v\rangle\neq 0$


Here the vector $v$ is nonnegative if it has nonnegative entries. The $\langle a,v\rangle$ denotes the standard scalar product. I am looking for the $a$'s whose lengths would polynomially depend on $n$ (actually my cost is the maximum of the absolute values $|\langle a,v\rangle|$ over all $v$'s as above).


The question is motivated by my attempts to find a convenient way to cancel out, in calculations, the inputs of nontrivial partitions, leaving only the one coming from $1+1+\ldots+1=n$. 



Separate the trivial partition by a linear hyperspace
There is no chance you can keep $a$ of polynomial size. Indeed, let $a$ be any vector with the sum of absolute values less than ${n\choose [n/2]}/2$. Then among the sums $\sum\_{k\in I}a\_k$ where $I\subset\{1,\dots,n\}$, $|I|=[n/2]$ there are two equal, meaning that there is a non-trivial vector $w$ consisting of $0,\pm1$ such that $\langle a,w\rangle=\langle e,w\rangle=0$. Then $v=e+w$ is your killer.


