Solution:

Let $n$ full-price tickets sell at $p$ dollars each. Then $np+(140-n)\dfrac p2=2001,$ so $p(n+140)=4002=2\cdot3\cdot23\cdot29.$

Since $140\le n+140\le280,$ the only factor of $4002$ in range is $174=2\cdot3\cdot29.$ So $n+140=174,$ giving $n=34$ and $p=23.$ The full-price tickets raise $34\cdot23=782$ dollars.

Thus, the correct answer is A.