14-2j-m1

For convenience, let $n \to x$ denote the move of erasing $n$ and writing $x$; by the rules, we must have $n > x \ge k$ and $(n, x) = 1$.

Claim A. If $m$ is a good number, and $n > m$ is coprime to $m$, then $n$ is a bad number.
Proof. Because Ana only needs to choose $n \to m$, and then copy Banana's winning strategy starting from $m$.

Claim B. Any two good numbers cannot be coprime.
Proof. This follows immediately from Claim A and B.

Claim 1. If $n$ is a good number and $n|n'$, then $n'$ is a good number.
Proof. If $n'$ is a bad number, this means Ana can make the move $n' \to x$ with $x$ being a good number. However, $(n', x) = 1 \Rightarrow (n, x) = 1$, but $n$ and $x$ are both good numbers, which contradicts Claim C.

Claim 2. If $rs$ is a bad number, then $r^2s$ is also a bad number.
Proof. $rs$ being a bad number means Ana can make the move $rs \to x$ with $x$ being a good number, but $x$ is clearly coprime to $r^2s$, so from $r^2s \to x$ we know $r^2s$ is a bad number.

Claim 3. If $p > k$ is a prime and $n \ge k$ is a bad number, then $np$ is also a bad number.
Proof. If not, then there exists a smallest bad number $n$ such that $np$ is a good number. We derive a contradiction as follows.
1. Since $n$ is a bad number, Ana can make the move $n \to x$, where $x$ is a good number. It is easy to see that $(np, x) > 1$, otherwise $np$ would be a bad number, a contradiction. But we know $(n, x) = 1$, so $p|x$. Let $x = p^r y$, where $(p, y) = 1$.
2. Note that $y = 1$ is impossible, because if $y = 1$, then $x = p^r$; and since $(p, k) = 1$, Ana can make the move $x \to k$, so $x$ would be a bad number, a contradiction. Hence $y > 1$, so there must exist a smallest positive integer $\alpha$ such that $y^\alpha \ge k$.
3. Based on the fact that $np$ and $y^\alpha$ are coprime and $np$ is a good number, by Claim B, $y^\alpha$ must be a bad number.
4. By the minimality of $\alpha$, we have $y^\alpha < ky < py = \frac{x}{p^{r-1}} < \frac{n}{p^{r-1}}$, so $p^{r-1}y^\alpha < n$. Hence, by the minimality of $n$, $p^{r-1}y^\alpha$ must be a good number (because $x = p(p^{r-1}y^\alpha)$ is a good number). By the same argument, $p^{r-2}y^\alpha, \dots, y^\alpha$ must also all be good numbers.
5. But $np$ and $y^\alpha$ are both good numbers, so by Claim B, $(np, y^\alpha) > 1$, which contradicts $(n, x) = 1$ and $(p, y) = 1$. This completes the proof.

Now let $P_k(x)$ denote the set of prime factors of $x$ that are less than or equal to $k$. In what follows, we call two numbers $a, b$ similar if and only if $P_k(a) = P_k(b)$. To prove the original statement, we only need to show: if $a, b$ are similar, then $a, b$ are both good or both bad simultaneously. Note that $ab$ is similar to both $a$ and $b$ at the same time, so this is equivalent to: if $c \ge k$ is similar to some multiple $d$ of it, then $c, d$ are both good or both bad simultaneously.

proof If not, then there exists a smallest $d_0$ such that there exists a factor $c_0 \ge k$ of it such that $c_0, d_0$ differ in being good or bad. By the Claims above, such a $d_0$ does not exist.