A Joule is a Newton-meter and is also a Watt-second. Gravity is about 9.81 Newtons/kilogram.
Raising 1 pound 1000 feet would be raising 0.4536kg 304.8 meters. So that would be 9.81 * 0.4536 * 304.8 = 1356 Joules, or 1356 Watt-seconds.
Your peak sustained energy output is probably in the general range of 300 watts (and "cruising" would be somewhere around half that), so you'd have to use all your energy for about 4.5 seconds to raise that one pound 1000 feet. (Or, to put it in perspective, about 19 minutes to raise a 250 pound bike+rider 1000 feet.)
For your assumed 250 watts, this would be 5.4 seconds for one pound or 22.6 minutes for 250 pounds. This would produce a speed, on the 10,000 foot distance, of about 5 mph. (Note that dropping much below roughly 200 Watts will produce a speed too slow to stay upright, especially given that the slower you go the more energy you must expend trying to stay upright.)
Of course, this is ignoring wind and rolling resistance losses, and hence the time required to "cover the distance" on level ground. Rolling resistance would be about the same as on level ground, but wind resistance would be less, since you're moving slower, and wind resistance is generally the larger of the two. So you need to add to the above times maybe 1/2 or 2/3rds of the time it would take you to cover the same distance on level ground. For a 10% grade that would be the time to cover 10,000 feet or about 1.9 miles. At 15 mph that would be about another 7.5 minutes, so add maybe half that.
Rats -- Just realized the question was in kg and meters...
"If I on one ride (of 10,000 meters at 10%) add 1 kg of weight to the bike, how much slower (in time) will I be (assuming 250 watts output)?":
That would be 9.81 * 1kg * 1000 meters * = 9810 watt-seconds. At 250 watts that's 38.84 seconds additional time due to adding 1kg.
And it occurs to me... that one could use the same calculations backwards in order to roughly calculate wattage output, given a weight, an average speed, and an average slope. This would likely be more accurate than many other wattage estimation schemes.