Is n/2^n always a finite sum of distinct terms a/2^a? (Erdős #261)

For a positive integer $n$, consider representations $$\frac{n}{2^n}=\sum_{k=1}^{t}\frac{a_k}{2^{a_k}}$$ with $t\ge 2$ and distinct positive integers $a_1,\ldots,a_t$. (1) Are there infinitely many $n$ admitting such a representation? (2) Does every $n$ admit one? (3) Is there a rational $x$ for which the infinite equation $x=\sum_{k=1}^{\infty}\frac{a_k}{2^{a_k}}$ (with distinct positive integers $a_k$) has at least $2^{\aleph_0}$ solutions — or, in Erdős's weaker form, at least two solutions?

Acceptance. FULLY RESOLVES (any one open part): (2) PROVE that every positive integer $n$ admits a representation $\frac{n}{2^n}=\sum_{k\le t}\frac{a_k}{2^{a_k}}$ with $t\ge2$ and distinct $a_k$ (a complete proof), OR EXHIBIT a specific $n$ with a machine-checkable proof that no such representation exists (a finite check, since any representation confines the $a_k$ to a bounded exponent window); (3) EXHIBIT a rational $x$ together with two explicitly described distinct infinite representations $x=\sum a_k/2^{a_k}$ (verifiable via a closed-form or eventually-periodic description and summation), settling Erdős's weaker two-solution form, or prove some rational has continuum-many. ADVANCES (checkable): extend the verified range for part (2) — confirm every $n\le N$ is representable for a new record $N$ strictly beyond the best bound stated in the background ($10^4$), delivering the search program and a per-$n$ certificate (an explicit representation for each $n$ in range); or prove representability for an infinite family strictly larger than the Borwein–Loring family. Deliver the proof, the explicit witness/counterexample, or the search code plus extended verified range with per-$n$ certificates.