The game with countable amount of steps Here is a cute problem.
The angel and the devil play a game. Firstly the angel has an empty box and the devil has a box which contains all numbers from $\mathbb{N}$ (one copy of every natural number). Then the game begins, players do their moves on turns.  
The angel puts $k$ natural numbers into his box (it should be new numbers which are not already in), the devil throw out one number from his box.
The devil wins if after countable number of turns
— his box contains infinite amount of naturals
and
— if the devil has a natural, the angel has it (in his box) too.  
"Countable number of turns" here could be described as it has done in Ross–Littlewood paradox 
Consider two cases: $k=1$ and $k=2$.
The question is (obviously): who will win and what is the correct strategy for the winner?
Well, I suppose I have an answer:

 If $k=1$ the angel wins, the key is to add the number the devil threw out. If it is impossible — just add the least one. 
 If $k=2$ the winner is the devil, the key is to throw out the least of all numbers angel does not have — because of $k=2$ it is infinite number of naturals to stay.


And slightly more difficult one (and I don't have an answer):
initial conditions are the same, the angel can put any finite (including $0$) number of naturals, the devil can throw out one or zero naturals.
The devil wins if after all turns are done the set $D$ of naturals in his box is infinite and either $D \cap A$ or $D \setminus A$ is finite, where $A$ is the set of naturals in the angel's box.
Who will win? What is the strategy?

Edited, thx to Henning Makholm.
 A: Your "slightly more difficult" game is a win for the devil. Here is a winning strategy for the devil: At each turn, he considers the set of all natural numbers $x$ such that (1) $x$ is still in the devil's box, and (2) $x$ is not in the angel's box, and (3) some natural number greater that $x$ is in the angel's box. If that set is nonempty, he throws out its least element; otherwise he does nothing.
At the end of the game, either $A$ is finite and $D$ is cofinite, or else $A$ is infinite and $D$ is an infinite subset of $A$.
P.S. The devil still wins if we make several changes in the rules, all in the angel's favor: (a) at each turn, the angel may put any finite or infinite set of natural numbers (including all or none of them) in his box; (b) at each turn the devil must throu out exactly one (new) number from his box; (c) the devil wins if $D$ is infinite and either $D\cap A$ is finite or $D\setminus A$ is empty. The devil's winning strategy is a slight modefication of the strategy described above, namely:
At each odd-numbered turn, the devil throws out the smallest odd number which is still in his box.
At each even-numbered turn, the devil considers the set of all even numbers $x$ such that (1) $x$ is still in the devil's box, and (2) $x$ is not in the angel's box, and (3) some even number greater than $x$ is in the angel's box. If that set is nonempty, he throws out its least element; otherwise he throws out the smallest odd number which is still in his box.
