My apologies; I was commenting on the thread, not the question which appeared to be answered already. However, if you insist:
There are four factors determining the weight of the helium cylinder:
1. The force of gravity where the weighing is taking place
2. The weight of the cylinder itself
3. The weight of the volume of air displaced by the cylinder
4. The weight of the contents of the cylinder (the helium inside)
Releasing or withdrawing helium from the cylinder would not significantly effect 1, 2, or 3 since any helium released into the air would quickly dissipate upwards (helium having greater buoyancy than air).
This leaves us with only the helium inside the cylinder to consider.
The weight of the helium is determined by its mass, this being the weight of a single helium atom times the quantity of helium atoms inside the cylinder. Therefore, as helium leaves the cylinder, the combined weight of the cylinder and its contents diminishes by the amount of helium released, unless . . .
If the helium from the cylinder is released within a confined enclosure of fixed size (such as a larger sealed tank) the helium (along with any air present) would be compressed until the pressure inside and outside the cylinder reached equilibrium. Under these circumstances the cylinder and the helium remaining inside would experience greater buoyancy and consequently an increased reduction in weight for the amount of helium released.
Either way the answer to your question is �No�. In fact, quite the opposite is true. A helium cylinder can only get lighter as it becomes empty.
Simply stated, removing mass (of any kind) from a given volume of space can only reduce the weight of that volume as measured under a specified force of gravity.