If I assume that the hole in the top equalizes the pressure in the tube with the ambient pressure at 100 m and the generator replenishes exactly the amount of helium that escapes at the top then I think it goed something like this.
1 atm at 0 m and 20°C = 101325 Pa
air density = 1.2 kg/m^3
P_air @ 100 m = 1 atm + 100 m * 1.2 kg /m^3 * g
P_tube @ 100 m = P_air @ 100 m = 100148 Pa
water vapour pressure @ 20°C = 2338 Pa
P_He = 1 atm - wvp = 98987 Pa
n_He = PV/RT = 40.6 / m^3 = 0.163 kg / m^3
n_H2O = 0.96 / m^3 = 0.017 kg / m^3
mixture density = 0.18 kg / m^3
P_tube @ 0 m = P_tube @ 100 m + 100 * 0.18 * g = 100325 Pa
dP @ 0 m = 1000 Pa, i.e. the water in the tube will be 100 mm higher.
You could do some iterations where you recalculate the densities using the average pressure at 50 m for example.
1 atm at 0 m and 20°C = 101325 Pa
air density = 1.2 kg/m^3
P_air @ 100 m = 1 atm + 100 m * 1.2 kg /m^3 * g
P_tube @ 100 m = P_air @ 100 m = 100148 Pa
water vapour pressure @ 20°C = 2338 Pa
P_He = 1 atm - wvp = 98987 Pa
n_He = PV/RT = 40.6 / m^3 = 0.163 kg / m^3
n_H2O = 0.96 / m^3 = 0.017 kg / m^3
mixture density = 0.18 kg / m^3
P_tube @ 0 m = P_tube @ 100 m + 100 * 0.18 * g = 100325 Pa
dP @ 0 m = 1000 Pa, i.e. the water in the tube will be 100 mm higher.
You could do some iterations where you recalculate the densities using the average pressure at 50 m for example.