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Source. Erdős (1945), Theorem 1 and its proof, printed p. 898 (published scan).

Statement. Let $N\ge1$ and let $x_1,\ldots,x_N$ be real numbers with $|x_i|\ge1$. In any open interval of length two, the number of assignments $\varepsilon\in\{-1,1\}^N$ satisfying $\sum_i\varepsilon_i x_i$ in that interval is at most

$$B_N=\binom N{\lfloor N/2\rfloor}.$$

The same bound holds for either half-open interval of length two. The bound is attained for every $N\ge1$.

Proof. Replace a negative $x_i$ by $-x_i$ and simultaneously replace its sign coordinate $\varepsilon_i$ by $-\varepsilon_i$. This is a bijection of assignments preserving every sum, so assume all $x_i\ge1$.

For an assignment let $A=\{i:\varepsilon_i=1\}$. Its sum is

$$Z_A=2\sum_{i\in A}x_i-\sum_{i=1}^N x_i.$$

If $A\subsetneq D$, then

$$Z_D-Z_A=2\sum_{i\in D\setminus A}x_i\ge2.$$

Any two points of an open or half-open interval of length two have distance strictly less than two. The subsets corresponding to the qualifying assignments therefore form an antichain. By Theorem 4 with $r=1$, this family has size at most $B_N$. Distinct assignments correspond to distinct subsets, so coincident numerical sums are counted with their proper multiplicity.

For sharpness, take all $x_i=1$. A rank-$k$ subset gives sum $2k-N$ with multiplicity $\binom Nk$. The open interval of length two centered at $2\lfloor N/2\rfloor-N$ contains exactly the central rank, since consecutive rank sums differ by two. Its count is $B_N$. $\square$

Scope. The source cites Sperner's theorem; the linked same-paper shadow proof supplies that input here. The source explicitly gives sharpness for even $N$; the displayed center also handles odd $N$. A closed interval of length two does not satisfy the theorem: for $N=1,x_1=1$, the interval $[-1,1]$ contains two assignments, whereas $B_1=1$.

Bears on. Problem 498: proves the problem's bound $B_N$ when every input is real, since an open unit disk meets the real line in an open interval of length at most two.