1 is a solution, because 1/1 = 1 and 1/1= 5*1-4

The next solution would need to sum to 6 and contain exactly 2 fractions, but only 1/2+1/2 contains 2 fractions totalling 1.

The next solution would need to sum to 11 and contain exactly 3 fractions:

1/2+1/3+1/6=1, and 2+3+6=11

The next solution would need to sum to 16 and contain exactly 4 fractions:

1/4+1/4+1/4+1/4 = 1, and 4+4+4+4=16

The next solution would need to sum to 21 and contain exactly 5 fractions. However, 21 does not have any representation as the sum of positive integers (not necessarily distinct) whose reciprocals sum to 1.

The next solution would need to sum to 26 and contain exactly 6 fractions. For example, 1/4+1/4+1/6+1/6+1/6=1, but that has only 5 fractions.

25 is a perfect square. Clearly for any such square x, one representation is 1/x+1/x…(x times) = 1. This representation of x in terms of the sum of positive integers (not necessarily distinct) whose reciprocals sum to 1, is maximal in the sense that it contains the most fractions for its size.

I haven't found an exhaustive set of solutions with 6 fractions, but by way of demonstration there are in all 147 solutions with x⩽y⩽z⩽a⩽b, x,y,z,a,b being 5 fractions whose reciprocals sum to 1.

We need only consider the last few:

137) [3, 4, 6, 6, 12] 31

138) [3, 4, 6, 8, 8] 29

139) [3, 5, 5, 5, 15] 33

140) [3, 5, 5, 6, 10] 29

141) [3, 6, 6, 6, 6] 27

142) [4, 4, 4, 5, 20] 37

143) [4, 4, 4, 6, 12] 30

144) [4, 4, 4, 8, 8] 28

145) [4, 4, 5, 5, 10] 28

146) [4, 4, 6, 6, 6] 26

147) [5, 5, 5, 5, 5] 25

The series (5n-4) then continues 31, 36, 41,... with n=7,8,9,... For n=8, say, we know already that the maximal representation in the above sense is 1/6+1/6+1/6+1/6+1/6, with a mere 6 terms.

I therefore conclude that the list first given above is exhaustive.