I like using one I only remember from my dad when I was a kid and I don't know where it came from, nor have I ever seen it used in popular media the way others have, like the two doors/liar puzzle or the weighted jug one.
It involves two gold ropes hanging from rings in the ceiling which is high enough to die from should you fall. The puzzle requires obtaining both lengths of rope without dying. I can't remember the whole thing off-hand (I have it written down, physically in a notebook; but it's still in a box buried in a closet right now). It's such a convoluted puzzle that if I hadn't written the whole thing, and the solution down, I can't remember it and would never be able to solve it if I ran into it.
Edit: I have dug out the notebook and found the riddle for y'all. I almost had it right in my head when I made the comment, but psyched myself out lol
There are two lengths of rope made of a magical gold alloy, attached by rings in the ceiling about 1 foot apart from each other. The ropes are pretty thick, but still able to be climbed without assistance. The rings are 108 feet from the ceiling, and there is a special knife on a nearby pedestal that can cut through the ropes without effort.
The goal is to try and obtain as much hold rope as possible; but how does one do it?
Answer:
The solution to get 100% of both ropes is to tie the ends of the ropes together at the bottom, climb up and cut 1 rope free of its ring. You then pull that rope through the ring of the remaining rope until the knot from the bottom ends that are tied together is just past the ring, grab the loose rope to support your weight while you cut the second rope and then shimmy down holding both sides of the rope, and once at the bottom, you just pull both ropes free of the ring. This is easier to demonstrate with a physical mock-up, which is pretty easy to do with some key rings, styrofoam blocks and twine.
I find it fun to use because even a failure can technically yield a reward, just not as much of one if the only reward is the rope and knife itself.