We shall prove that the only possible integers are perfect squares. We first prove that if $N=n^2$ then $A$ always wins. The numbers on the blackboard will look like this: $1, d_1,\dots ,d_k, n, g_1,\dots ,g_k, n^2$ Where $g_i=\frac{n^2}{d_i}$. It's clear that each $d_i$ is a divisor of $n$ and because of this we have $d_i | g_i$ $\forall i$. Now, after $A$'s first move, partition the numbers on the blackboard into $k+1$ couples as follows: $(1,n); (d_1, g_1); \dots ;(d_k, g_k)$. For every move of $B$, $A$ will erase the remaining number in the chosen couple. Hence, $A$ wins. If we have a non-square number, we can prove by induction on the number of primes $k$ where $n=p_1^{\alpha_1}\cdots p_k^{\alpha_k}$ that $B$ can always win if the divisors of $n$ is an even number (it happens iff $n$ is non-square). We can prove this for $n=1$. Suppose it's true for $k-1$. Then $n=g\cdot p_k^{\alpha_k}$. We can partition the divisors as follows: $A_0=${divisors of $g$}, $A_1=${divisors of $g$ multiplied to $p_k$},$\dots$, $A_{\alpha_k}=${divisors of $g$ multiplied to $p_k^{\alpha_k}$}, . It's clear that if $B$ has a strategy for $A_0$ then he also has it for $A_i$ $\forall i$.

Fail. $4 \cdot 9=36$, but 9 isn't multiply of 4.

Swistak wrote: Fail. $4 \cdot 9=36$, but 9 isn't multiply of 4. You're right, that's an awful mistake. I think though that we can prove it by induction in the same way as in the second part (as long as the second part is right).

now if n is a perfect square then we shall prove that we can pair all divisors of n except n into pairs such that in every pair one number divides the other one with induction on how many different prime divisors n has (k): base: if k=1 $n={p_1}^{a_1}$ we pair $(1,p_1)$; $({p_1}^2,{p_1}^3)$;...;$({p_1}^{a_1-2},{p_1}^{a_1-1})$ which is ok because $a_1$ is divisible by 2 assumption: we can pair the divisors of any m which has k-1 different prime divisors except m in a way described above step: $ n=p_{1}^{a_{1}}\cdots p_{k}^{a_{k}} $ we use assumption to pair the divisors of $ m=p_{1}^{a_{1}}\cdots p_{k-1}^{a_{k-1}} $ for every paired pair of numbers x and y we pair numbers ${p_k}^i x$ and ${p_k}^j y$ for every $i=1...a_k$ and that way we have paired every divisor of n except those which are multipliers of m and them we pair: we pair $(m,mp_k)$; $(m{p_k}^2,m{p_k}^3)$;...;$(m{p_k}^{a_k-2},m{p_1}^{a_k-1})$ of course using that $a_k$ is even. Now we have paired everything except n as we desired Now after As first move A always plays the pair of the number which B played so A can always play its move if B was able to play its before it so clearly A wins. If n is not a perfect square than if $ n=p_{1}^{a_{1}}\cdots p_{k}^{a_{k}} $ than there must be some odd $a_i$ WLOG assume i=k let b be the number of divisors of $ m=p_{1}^{a_{1}}\cdots p_{k-1}^{a_{k-1}} $ and lets call that divisors $d_1<d_2<...<d_b$ now for every $i=1...b$ we pair $(d_i,d_i p_k)$;$(d_i {p_k}^2,d_i {p_k}^3)$;...; $(d_i{p_k}^{a_k-1},d_i{p_k^}^{a_k})$ now note that we have paired every positive divisor of n (including n) such that in every pair one number is divisor of the other one so B can chose a pair of any number A have chosen before him and can always play a move after A-s move so B wins and we are done.

nice proof. You can prove easier that A loses if $n$ isn't perfect square. If $n$ isn't square, then $n$ has even number of divisors. Let $m$ be the number which A plays. B can always choose number $n$ / $m$, and so B can erase the last number on the table and A loses.

If n=6 and A choose 2, B can't choose $\frac{6}{2}=3$ !!