A gap-at-most-two representation of $1$ must have at least two gaps equal to $1$

Let $1 < n_0 < n_1 < \cdots < n_{k-1}$ with $k \ge 2$ and $\sum_i 1/n_i = 1$, and suppose every consecutive gap satisfies $n_{i+1} - n_i \le 2$ — that is, suppose the representation is a counterexample to Erdős problem #287. Then at least two of the gaps are equal to $1$.

A gap-$\le 2$ sequence breaks into maximal runs of step exactly $2$, and consecutive runs have opposite parity, so a unit gap is precisely a parity switch. The theorem therefore says that a counterexample must contain at least three such runs, and in particular at least two blocks of even denominators.

The proof is a two-scale $2$-adic argument. With at most one unit gap the denominators consist of at most two runs of step $2$, one of which is even, say $2m, 2(m+1), \dots, 2(m+t-1)$; its reciprocal sum is $\tfrac12 \sum_{j<t} 1/(m+j)$. By Kürschák's lemma a block of consecutive integers has a unique element of maximal $2$-adic valuation, so $\left|\sum_{j<t} 1/(m+j)\right|_2 \ge 1$ and hence the even run contributes $2$-adic norm $\ge 2 > 1$, while the remaining, all-odd, run contributes norm $\le 1$. The total therefore has $2$-adic norm $> 1$ and cannot equal $1$. The degenerate case in which every denominator is odd is excluded separately: there all gaps equal $2$, so the denominators are every odd number in their range, and by Bertrand's postulate that range contains a prime past its midpoint — which cannot be a denominator.

Together with the classical result that the gaps cannot all equal $1$, this narrows any hypothetical counterexample to a representation whose gap sequence mixes at least two $1$s with the $2$s.