TITLE: Reverse Littlewood-Offord problem: lower bound for the number of choices of signs such that $|\pm a_1\dots\pm a_n| \leq \max|a_i|.$
QUESTION [10 upvotes]: Let $n$ be a positive integer. For $\mathbf{a}\in\mathbb R^n$ let $N(\mathbf{a})$ be the number of choices of $\epsilon\in\{-1,1\}^n$ such that
$$\Biggl|\sum_{i=1}^n\epsilon_ia_i\Biggr|\leq \max_{1\leq i\leq n}|a_i|.$$
Example. for $n$ even and $a=(1,\dots,1),$ we are counting $\epsilon$ with $|\sum_{i=1}^n\epsilon_i|\leq 1,$ which forces $\epsilon$ to have exactly $n/2$ entries of $-1,$ so $N(1,\dots,1)$ is the central binomial coefficient $\binom{n}{n/2}.$

Is it true that $N(\mathbf a)\geq \binom{n}{n/2}$ for even $n$?

I also have a conjectured lower bound for odd $n$ if you're interested. Define $$B_n=N(1,\dots,1,\tfrac12)=2\binom{n-1}{\lfloor (n-1)/2\rfloor}=
\begin{cases}
\binom{n}{n/2}&\text{ for $n$ even}\\
2\binom{n-1}{(n-1)/2}&\text{ for $n$ odd.}
\end{cases}
$$
I've confirmed $N(\mathbf a)\geq B_n$ by picking random vectors $\mathbf{a}$ uniformly at random, but these tests aren't very convincing since it doesn't even find $B_n$ for $n\geq 9.$
For $n=3$ we always have $N(\mathbf{a})\geq B_3=4$: assume $1=a_1\geq a_2\geq a_3\geq 0,$ then $a_1-a_2-a_3$ and $a_1-a_2+a_3$ and their negations all lie in $[-1,1].$
This would be a reversed form of the Littlewood-Offord problem, where Erdős showed there are at most $\binom{n}{\lfloor n/2\rfloor}$ vectors $\epsilon\in\{-1,1\}^n$ such that $\Bigl|\sum_{i=1}^n\epsilon_ia_i\Bigr|\leq \color{red}{\min}_{1\leq i\leq n}|a_i|.$
It's a version of Tomaszewski's Problem with the $\ell^2$ norm replaced by $\ell^\infty.$
I believe a positive answer would also provide a lower bound for the Minimum number of balanced partitions.

REPLY [4 votes]: Here is a proof of a strong version of the conjecture in @mathworker21's post, which in the context of the original question proves $N(\mathbf{a})\geq B_n.$ This also proves the slightly stronger "balanced partitions" conjecture by Karo that I linked to. (That post only asked about even $n,$ so I posted a cut-down version of this argument there.)

$\textbf{Theorem}$: Let $\Sigma$ be a collection of subsets of $[n]$ satisfying

$A \in \Sigma \implies A^c \in \Sigma$;
$\Sigma$ is "convex" i.e. $A \subseteq B \subseteq C$, $A,C \in \Sigma \implies B\in \Sigma$;
$\Sigma$ is a "cutset" i.e. $\emptyset = A_0 \subseteq A_1 \subseteq \dots \subseteq A_{n-1} \subseteq A_n = [n]$ with $|A_j| = j$ for each $j$ implies $A_j \in \Sigma$ for some $j$.

Then $|\Sigma| \ge B_n$.

Proof.
Define $L$ to be the set of sets $A\subseteq[n]$ such that $A\not\in\Sigma$ and there exists $B\in\Sigma$ with $A\subset B.$
Define $U$ to be the set of sets $A\subseteq [n]$ such that $A\not\in\Sigma$ and there exists $B\in\Sigma$ with $B\subset A.$
By the cutset property, every set is in $L\cup\Sigma\cup U.$ And by convexity, $L\cap U=\emptyset.$ So every subset of $[n]$ is in exactly one of $L,\Sigma,$ or $U.$
The L stands for lower, and the U stands for upper, in the sense of lower and upper sets.
By property 1, a set is in $U$ iff its complement is in $L.$ In particular, $|L|=|U|.$
By the cutset property, for any $A\in L$ we have $A\cup\{j\}\not\in U$ for any $j\not\in A.$ And since $L$ is a lower set, $A\setminus\{j\}\not\in U$ for any $j\in A.$ This means that $L$ and $U$ are at Hamming distance at least $2$:
$$d(L,U):=\min_{A\in L}\min_{B\in U}|A\Delta B|\geq 2.$$
By the result in https://cseweb.ucsd.edu/~ccalabro/essays/harper.pdf, or B. Bollobás, Combinatorics: Set Systems, Hypergraphs, Families of Vectors, and Combinatorial Probability Chapter 16 Theorem 3, there is an integer $r$ and collections $L_0,U_0$ of subsets of $[n]$ such that:

$|L_0|=|U_0|=|L|=|U|$
$d(L_0,U_0)\geq 2$
all sets in $L_0$ have cardinality at most $r$
$L_0$ contains all sets of cardinality less than $r$
all sets in $U_0$ have cardinality at least $n-r$
$U_0$ contains all sets of cardinality more than $n-r$

Even $n.$
Suppose for contradiction that $L_0$ contains any set $A$ of cardinality $n/2.$ Then $r\geq n/2,$ so $U_0$ contains all sets of cardinality $(n/2)+1.$ In particular, $A\cup\{j\}\in U_0$ for any $j\in[n]\setminus A.$ This contradicts $d(L_0,U_0)\geq 2.$
So $L_0$ does not contain any set of cardinality $n/2.$ By a symmetric argument, neither does $U_0.$ This implies
$$|\Sigma|=2^n-|L|-|U|=2^n-|L_0|-|U_0|\geq \binom{n}{n/2}=B_n.$$
Odd $n.$
By a similar argument to the even case, $L_0$ consists of sets of cardinality at most $(n-1)/2,$ and $U_0$ consists of sets of cardinality at least $(n+1)/2.$
Let $k=(n+1)/2,$ so $n=2k-1.$
Let $L'_0$ denote $\{A\in L_0\mid |A|=k-1\}.$
Let $U'_0$ denote $\{A\in U_0\mid |A|=k\}.$
Let $\Delta(U'_0)$ denote the lower shadow of $U'_0,$ i.e. the sets $B$ with $|B|=k-1$ and $B\cup\{j\}\in U'_0$ for some $j.$
Suppose for contradiction that $|L'_0|> \binom{2k-2}{k}.$
The Kruskal-Katona theorem says
$$|L'_0|=|U'_0|\geq\binom{2k-2}{k}\qquad\implies\qquad|\Delta(U'_0)|\geq \binom{2k-2}{k-1}.$$
But $\binom{2k-2}{k}+\binom{2k-2}{k-1}=\binom{2k-1}{k}=\binom{2k-1}{k-1}=\binom{n}{k-1}.$ By the pigeonhole principle applied to the sets of cardinality $k-1,$ the sets $L'_0$ and $\Delta(U'_0)$ would have to intersect, which contradicts $d(L_0,U_0)\geq 2.$
So $|L'_0|\leq\binom{2k-2}{k}.$ The cardinality of $\Sigma$ is the number of sets not in $L_0$ or $U_0,$ which is twice the number of $k-1$ element sets not in $L_0'.$ This gives $|\Sigma|=2(\binom{n}{k-1}-|L'_0|)\geq 2\binom{2k-2}{k-1}=B_n.$