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Source. Erdős (1945), unnumbered corollary, printed p. 899 (published scan). The printed strict inequality is corrected below.

Statement. Let $N\ge1$, let $r\ge1$ be an integer, and let $x_1,\ldots,x_N$ be real with $|x_i|\ge1$. The number of assignments whose signed sum belongs to an open interval of length $2r$ is at most

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

Proof. Write the target interval as $(u,u+2r)$. It is contained in the disjoint union

$$\bigcup_{j=0}^{r-1}[u+2j,u+2j+2).$$

Each piece is a half-open interval of length two and hence contains at most $B_N$ assignments by Theorem 1. Summing over the $r$ pieces proves the bound. The possible extra endpoint $u$ can only enlarge the counted set; every internal division point is included in exactly one piece. $\square$

Printed endpoint correction. The source says the count is less than $rB_N$. For $r=1$, even $N\ge2$ and $x_1=\cdots=x_N=1$, exactly $B_N$ assignments have sum in $(-1,1)$. The weak inequality is therefore necessary. This is a correction supplied by the compilation, not a cited author erratum. Radius zero and negative integers are not part of the geometric statement.

The sharper Theorem 3 replaces this coarse multiple by the sum of the $r$ largest binomial coefficients, truncated when $r>N+1$.

Bears on. Problem 498: an input to the order bound of Theorem 2, not the problem's exact bound.