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Source. Erdős (1945), closing discussion, printed pp. 901–902 (published scan). The paper states that the real case is easy. The full deduction follows.

Statement. Let $N\ge1$ and let $x_1,\ldots,x_N$ be real with $|x_i|\ge1$. For any $w\in\mathbb C$, let $Q_{\mathrm{in}}$ count the sign assignments with $|Z-w|<1$ and let $Q_{\mathrm{bd}}$ count those with $|Z-w|=1$, where $Z=\sum_i\varepsilon_i x_i$. Then

$$Q_{\mathrm{in}}+\frac12Q_{\mathrm{bd}}\le B_N.$$

Proof. First suppose $w=u$ is real. By the half-open version of Theorem 1, each of the intervals $[u-1,u+1)$ and $(u-1,u+1]$ contains at most $B_N$ assignments. Averaging their counts gives weight one to each interior assignment and weight one half to each endpoint assignment. This is exactly $Q_{\mathrm{in}}+\tfrac12Q_{\mathrm{bd}}$.

Now write $w=u+iv$ with $v\ne0$. Every signed sum $Z$ is real. If $|v|>1$, no such sum satisfies $|Z-w|\le1$, and the assertion is immediate. If $0<|v|\le1$, all qualifying sums lie in

$$\left[u-\sqrt{1-v^2},\;u+\sqrt{1-v^2}\right].$$

This is a closed interval of length strictly less than two; at $|v|=1$ it is a singleton. It is contained in the open interval $(u-1,u+1)$, so its entire unweighted assignment count is at most $B_N$ by Theorem 1. The weighted count is no larger. $\square$

Scope. This proves the real-input assertion only. The corresponding complex-input statement is one of the paper's dated conjectures. The theorem is not a bound for the unweighted count in an arbitrary closed interval of length two.

Bears on. Problem 498: the real-input case of the weighted strengthening the paper proposes for the problem's bound.