A riddle on positive integers 
Two boys A and B tell the teacher a positive integer each, but neither of them know the other's number. The teacher writes two positive integers on the blackboard and declares that one of them is the sum of the two numbers.
Then he asks A, "can you guess the sum of the two numbers?" If the answer is "no" then the teacher asks B the same question. Suppose the boys are truthful and intelligent. Prove that one of the answers will eventually be "yes".

I am absolutely clueless about where to start. I tried taking some examples to see if the answer actually comes yes, but it seems that the boys are more intelligent than me!
 A: A useful way to think of the problem is to imagine the two boys aren't told their numbers, but instead are each given a box with some number of coins in it, told to privately count their coins, and then the teacher writes two (positive) numbers on the board, one of which is the correct total number of coins between the two boxes.  Furthermore, each time a boy answers "No," the teacher takes a coin from that boy's box, and, correspondingly, decrements the two written numbers each by one (since everyone knows that a coin has been removed).
It's now clear this cannot go on forever. In particular, as soon as one boy's box is empty and the two written numbers are, say $M$ and $N$, with $M\lt N$, that boy knows the other boy will say "Yes" if his box has $N$ coins in it (because he'll know the total cannot be less than the number of coins in his box), so if he says "No," his box must have $M$ coins in it, after which the boy with the empty box will be able to say "Yes" on his next turn.
A: Let A be given $a$ and B be given $b$.  Let the numbers on the board be $x,y$ with $x \lt y$  Even before any questions we know that $a,b \lt y$  
When A says no the first time, we are told $a \lt x$ because if it were greater or equal A would know the sum was $y$.  
When B says no the first time we are told $b \lt x$ for the same reason as above.  We are also told $b \gt y-x$ because if it were smaller B would know the sum was $x$.  
Now when A says no the second time we know $a \lt 2x-y$ because otherwise A would know the sum was greater than $x$ and it would have to be $y$  
As we keep going the ranges will be reduced by $y-x$ each time.  Eventually the range will become less than zero and the sum will be known.
A: Let's call $A,B,S,X$ respectively the numbers told by boy A, by boy B, the relevant sum, and the other number written by the teacher.
It is clear that the problem does not change if we add to or subtract the same amount from $B,S,X$ or from $A,S,X$.
Thus we can assume that the minimum of the four numbers is $0$.
a)  if the sum $S$ is $0$
   boy A  cannot decide between $S$ and $X$ , but his indecision will induce B to provide the correct answer.
b)  if $X$ is $0$
  boy A can decide for $S$.
c)  if $A$ is $0$
  boy A  cannot decide between $S$ and $X$ , but his indecision will induce B to provide the correct answer.
d)  if $B$ is $0$
  boy A  can decide  if $X<S$, otherwise his indecision will induce B to provide the correct answer.
