Banana cart problem.

Bob travels 500 miles with a 1,000 bananas, drops off 500, then returns 500 miles for another load of 1,000, three times. 1500 at midway total. He loads 1,000, brings them 250 miles, dropping off 750, returns to midway, loads 500 to new midway, having 1,000 then, 250 miles away, then arrives at city B with 750 bananas.

Bob can deliver 3,000 delicious, pre-digested bananas.

Start with an initial load of 1000, stop at mile 333 and stash 334, consuming the remaining 333 of the load on the return trip.
Start with a second load of 1000, pick up 333 from the stash at mile 333, continue to mile 500 and stash 334. Consume the remainder of the load plus the one left in the mile 333 stash on the trip back.
Start with a final load of 1000, pick up the 334 stashed at mile 500, deliver 334.

I think it would be best to cache 2000 bananas at the first stopping point, since this will leave him with only 2 trips to do on the next leg. He will eat 1000 bananas getting there, and it will take 5 one way trips to get there if he does three total trips to the first stop. The distance to stop point 1 is 200 miles.

Take 1000 bananas 200 miles.
Return with 200 bananas, which will be depleted upon return to city A.
Stop point 1 has 600 bananas.
Take 1000 bananas 200 miles.
Return with 200 bananas.
Stop point 1 has 1200 bananas.
Take 1000 bananas 200 miles.
Stop point 1 has 2000 bananas.

The second stop point should have 1000 bananas remaining. He will eat 1000 bananas getting there, and it will take 3 one way trips.
Take 1000 bananas 333 miles (just avoiding fractional bananas)
Return to stop point 1 with 333 bananas.
Stop point 2 now has 334 bananas.
Take the remaining 1000 bananas 333 miles.
Stop point 2 now has 1001 bananas.

Leave 1 banana behind and drive 1000 bananas the remaining 467 miles.
He will deliver 533 bananas to city B.

Take 1000 bananas 200 miles, drop off 600, go back to the start. Take another 1000 bananas 200 miles, drop off 600 and go back to the start. Take your final 1000 bananas, go 200 miles. You now have 2000 bananas 800 miles away. Take 1000 bananas and go 334 miles, drop off 333 and go back to the banana supply. Take 1000 bananas and go 334 miles. You now have 999 bananas and are 466 miles away. Keep walking and arrive at the market with (999 – 466) bananas, which is 533 bananas to sell. If you work with fractional miles, you may be able to make it there with 534, but I'm not sure

It seems obvious that he should set out with a full basket and therefore move 3000 in 3 loads of 1000 to a point where he then has 2000. He repeats this to get a stash of 1000 bananas; and finally just takes those to the end. Rounding errors means it's slightly different.

I am assuming he keeps an oddometer and starts to eat the banana just after the mile starts and finishes it just at the end - thus most trips he has to set out with 1000 bananas but after a mile has 999 - i.e. he cannot eat a banana and take 1000 bananas.

(i) To a point at Mile+200 takes five trips dropping off 600 (needs 200 to get there and 200 to get back), 600 (ditto) and 800 (only needs 200 to get there).
(ii) To a point 533 - trip 1 uses up 333 getting there, 333 getting back, so can drop off 334 - trip 2 uses up 333 so still has 667. Thus he has 1001 bananas.
(iii) It's best to ignore the 1 banana (as it costs more than that to pick it up) so makes for the end.

533 bananas make it to the end.

An alternative way to think of it is assume the driver goes backwards and forward by 1 mile increments.
Mile 0 to Mile 1 to 0 to 1 to 0 to 1 costs 5 miles, so 2995 bananas are at Mile1. Essentially every mile costs 5 bananas. Thus 2000 bananas get to Mile 200.
Similarly 200->201->200->201 costs 3 bananas - so every mile now costs 3 bananas - until mile 533.
At this stage, with 1001 bananas, the last trip (leaves one banana behind) is 467 miles, costs 1 banana per mile, so 533 bananas are delivered.

...the return trip after delivering his bananas? Does Bob need to bring a full load of bananas home to eat after making his delivery? If so, are there other bananas in City B to fuel his return?

Also, does he eat the banana before or after traveling the mile?

And finally, does reaching the border of City B at the end of the 1000th mile mark a successful delivery, or does he need to travel an additional distance to "cross the line"

Bob loads 1,000 banana's and drives off. He gets to city B, where he delivers his load, and then eats 1,000 bananas from their surplus in one sitting.

He loads his cart with 1,000 City B banana's and drives back to City A, eating one banana per mile so that when he returns, his cart is empty. Load the second 1,000 and head back, not eating until he gets to City B... repeat one more time, so that all 3,000 City A bananas are "delivered".

No change for the initial cached point:
Bob makes two round trips from City A to the 200 mile mark with 1000 banana loads, allowing him to cache 1200 bananas there while consuming 800.
Bob makes a final trip from City A with 1000 bananas to the 200 mile mark, consuming 200 and arriving with 800.
Bob has 2000 bananas at the 200 mile mark and no need of Ex-Lax.

Slight change for the second cache point:
Bob makes a 333.5 round trip from the 200 mile mark with 1000 bananas to the 533.5 mile mark, leaving 333 bananas there and consuming 667.
Bob takes the remaining 1000 bananas and heads for City B.
When he reaches the 533.5 mile mark he has consumed enough bananas to pick up the entire cache of 333 and continue on, assuming a banana in the process of being eaten is not part of the load.
For the final 800 mile leg, Bob consumes 800 bananas of 1333 and delivers the remaining 533 bananas.

Bob flies the banana stand three times to city B, never has any cravings delivering all 3000 bananas.

The same method is used to get 1001 bananas to Mile 533. There you have 1001 bananas - so eat one and take the other 1000. You arrive at Mile 534 with 1000 bananas on the cart and 466 miles to go; so eat a banana at miles 534 thru 999 (total eaten en route = 466) and deliver 534.

533 1/3 to city B.

Bob contacted his cousin Pedro, who has a small jet (for other cargo purposes) with a 250mph speed. That would take about a 4-hr flight from City A to City B. However, due to a medical condition, Pedro's banana intake is limited to 1 per hour. So Bob would still eat the pre-requisite bananas, for a total of 4 (1 banana for each 1 hr leg of the flight). The jet accommodates both Pedro's cart with the 1,000 bananas. Luckily, Pedro has other containers that take the remaining 2,000 bananas in City A.

Bob ties the two carts together and can get 2000 bananas to 333 1/3, and then gets 1333 bananas to the city using the same idea.

Bob never returns to his home, family and business in City A due to the severe banana shortage in City B? hehe....