It's small! Only 20 feet in diameter. Maybe 6 feet height at best. So 6x20 diameter in an "worst case" scenario, as if it were squared off on the edges, which it is not of course. It's sloping sides reduce volume considerably.
Using a lift of .078 pounds per cubic foot of helium at sea level (less at Ft. Collins' 5000 feet) a perfect 6 foot high, 20 foot diameter container would hold 1884 cubic feet. The lift would then be 146 pounds at sea level if fully inflated and the edgeds perfectly squared off. But reduce this to the saucer shape we've seen and you probably max out the volume at more like half that, or about 70 pounds lift at sea level. It would simply take too much pressure to push out the Mylar walls of the balloon much further without a rupture. At 5000 feet the atmospheric pressure is 83% of sea level. So multiply 70lbs by .83 and you get a lift of maybe 58 pounds or so. That's maybe a little more than the weight of a 6 year old boy. But you have to add in the weight of the balloon and the supposed "basket" as well. Clearly these would weight at least 15-20 pounds at the very least. That makes it unlikely.
Reference source for helium's lift at Yahoo answers. A cubic foot of helium can lift, the weight of the air it displaces. http://answers.yahoo.com/question/index?qid=20070107183...
"At sea level, air weighs about 0.078 pounds per cubic foot, so the upward buoyant force on a cubic foot of helium is about 0.078 pounds."
Info on barometric pressure at 5000 feet vs sea level at