77 is the greatest integer that cannot be a finite sum of distinct integers greater than 1 whose sum of their reciprocals is 1 In 1963, Ron Graham proved on a short article entitled "A Theorem on Partitions" (it can be found in the web) that every positive integer greater than $ 77 $ is a finite sum of distinct integers greater than 1 such that the sum of their reciprocals equals 1. 
Related: Prove that $ x_1+ \dotsb + x_k=n, \frac1{x_1}+ \dotsb + \frac1{x_k}=1$.
At the end of the article he says that D. H. Lehmer proved that $ 77 $ doesn't have this property, but hadn't published the proof at that time. On another Ron's article of 2008 or later he mention that result again and cite it as "D. H. Lehmer, (personal communication)".
Does anyone know how to prove it or where can I find a proof?
Thanks. Sorry for my english. 
 A: Greg Egan has verified, using Mathematica, that $77$ cannot be written as the sum of distinct positive integers whose reciprocals sum to $1$:
https://twitter.com/gregeganSF/status/1156839799610675200.
Christopher D. Long proved it using Sage; his code is here:
https://github.com/octonion/puzzles/tree/master/twitter/lehmer.
Dima Pasechnik wrote a very elegant Sage program that proves this result:
sage: from builtins import sum as psum
sage: def sumreseq1(l):
....:     return 1==psum(map(lambda t: 1/t, l)) 
sage: filter(sumreseq1, Partitions(11,max_slope=-1))
[[6, 3, 2]]
sage: filter(sumreseq1, Partitions(77,max_slope=-1))
[]

There are $58499$ ways to write $77$ as a sum of distinct positive integers, and one only needs to check these to prove the result.
Here, by the way, is the complete list of positive integers that cannot be written as the sum of distinct positive integers whose reciprocals sum to 1:

$2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 25, 26, 27, 28, 29, 33, 34, 35, 36, 39, 40, 41, 42, 44, 46, 47, 48, 49, 51, 56, 58, 63, 68, 70, 72, 77$

This is $A051882$ in the Online Encyclopedia of Integer Sequences.  However, I don't believe there is a reference to a proof.  Luckily, Mr. Drake has written a program that verifies this list.
A: Here's an attempt at a human-readable proof. Let such a partition be $X = \{ x_i \}, 1 \leq i \leq n$. Note that by modular considerations there cannot be a prime $p$ that only divides a single $x_i$. In particular, this implies that the largest such prime must satisfy $p + 2\cdot p <= 77$, and hence $p \leq 23$. But more is true; if $ Y = \{ p\cdot y_i \}, 1 \leq i \leq m$, are the terms of $X$ divisible by $p$, then $ \sum_{i=1}^{m} y/y_i \equiv 0 \pmod p$, where $ y = \prod_{i=1}^{m} y_i$. This rapidly eliminates $p > 11$.
Now $p=11$ is only slightly harder; the only candidate for $Y$ is $\{ 1\cdot 11,2\cdot 11,3\cdot 11 \}$, but this only leaves $77-6\cdot 11 = 11$ for the sum of the remaining terms of our partition. Now, the remaining terms can only be products of $2$ and $3$, leaving the only candidate for the remaining terms $\{ 2,3,6 \}$, but $1/2+1/3+1/6 = 1$ is too big. Thus the largest possible prime factor dividing an $x_i$ is 7.
The above argument also works for prime powers, showing that $3^2$ is the largest squared prime factor that may occur and any higher power must be a power of $2$. With some more work, you get a list of candidate $Y$ if $7$ divides some $x_i$ e.g. $Y = \{ 7, 42 \}, \{ 14, 35 \}, \{ 21, 28 \}$ or $Y = \{ 7, 14, 28\}$.
This should reduce the possible $X$ to a list checkable by hand.
