If sum of reciprocals of numbers is an integer, then there exists a subset with sum of reciprocals of 1

How about $\frac 13+4\cdot \frac 15+6\cdot \frac 17+\frac 1{105}=2$?

If we put them over a common denominator, it is $105$, so we need to find a subcollection of $1,21,21,21,21,15,15,15,15,15,15,35$ that sums to $105$. If we take all the ones ending in $1$ to get a multiple of $5$ we have $85$ and can't complete it. If we exclude all the ones ending in $1$ we are stuck again.

For a smaller list allowing repetitions, there is $\frac 12+2\cdot \frac 13+4\cdot \frac 15+\frac 1{30}=2$ I think eight terms will be hard to beat.

Added: We can find a set with no repetition. I find the reciprocals of $2,3,4,5,7,8,9,11,13,16,144,2574,30030$ add to $2$ and no subset of the reciprocals add to $1$.

This is an expansion of @Michael Lugo's comment.

The proposition is not true even if all denominators have to be unique. A counterexample: $(2,3,4,5,7,8,9,10,12,14,16,49,35280)$.

Here is a Python program which checks that these numbers are different, and the sum of the reciprocals of these numbers is 2, and that there is no subset of these numbers, where the sum of the reciprocals of the numbers in the subset is 1:

import fractions, itertools
R = fractions.Fraction
D = (2,3,4,5,7,8,9,10,12,14,16,49,35280)
assert len(D) == len(set(D))  # All elements are different.
assert sum(R(1,d) for d in D) == 2  # Sum of reciprocals is 2.
for s in xrange(len(D) + 1):
  for e in itertools.combinations(D, s):
    assert sum(R(1,d) for d in e) != 1  # Sum of reciprocals in subset is not 1.