Let $a$ and $b$ be the integers thought of by A and B, and let $N$ and $n$ be the integers given by the teacher, where $N>n$. Assume for contradiction that both A and B keep on saying "no" forever.
If $a\geq n$, then A knows $n$ cannot be the sum and will say "yes" the first time. Thus after A's first "no", B knows $a<n$.
If $b\leq N-n$, then $a+b<N$ (since $a<n$), so B knows $N$ cannot be the sum and will say "yes" the first time. Thus after B's first "no", A knows $b>N-n$.
If $a\geq 2n-N$, then $a+b>n$ (since $b>N-n$), so A knows $n$ cannot be the sum and A will say "yes" the second time. Thus after A's second "no", B knows $a<2n-N$.
If $b\leq 2N-2n$, then $a+b<N$ (since $a<2n-N$), so B knows $N$ cannot be the sum and will say "yes" the second time. Thus after B's second "no", A knows $b>2N-2n$.
If $a\geq 3n-2N$, then $a+b>n$ (since $b>2N-2n$), so A knows $n$ cannot be the sum and will say "yes" the third time. Thus after A's third "no", B knows $a<3n-2N$.
If $b\leq 3N-3n$, then $a+b<N$ (since $a<3n-2N$), so B knows $N$ cannot be the sum and will say "yes" the third time. Thus after B's third "no", A knows $b>3N-3n$.
By induction, this argument can be continued forever:
after B's $k$th "no", A always knows $b>k(N-n)$.
But $b$ cannot be greater than $k(N-n)$ for ALL $k$, so eventually B must say "yes" (unless A does first) and the game is up.
QED.
