We will prove that if $p$ is the smallest prime divisor of $k$, then $p-1$ is the smallest $k$-pable number. This implies the desired result, as it then follows that $\operatorname{ggd}(p-1, k)=1$. We know that $1+3+5+\ldots+(2 k-1)=k^{2}$. If $\ell$ is a $k$-pable number and $s$ is the sum of the elements in $B$, then the sum of the elements in $A$ is $\ell s$, so the total sum is $(\ell+1) s$ and this must equal $k^{2}$. Thus, $\ell+1 \mid k^{2}$. Since $\ell+1 \geq 2$, it follows that $\ell+1 \geq p$, where $p$ is the smallest prime divisor of $k$. Therefore, $\ell \geq p-1$. We will now prove that $\ell=p-1$ is a $k$-pable number, which immediately implies that it is the smallest $k$-pable number. And we are done, because $\operatorname{ggd}(k, p-1)=1$ since there are no prime divisors smaller than $p$ in $k$.
First, assume that $k$ is even, so $p=2$. We need to prove that we can partition the set $\{1,3, \ldots, 2 k-1\}$ into two sets with equal sums of the elements. We prove this by induction on $k$. For $k=4$ and $k=6$, we have $\{1,7\},\{3,5\}$ and $\{1,3,5,9\},\{7,11\}$, respectively. If we can partition the set $\{1,3, \ldots, 2 k-1\}$ in this way, we can also do it for the set $\{1,3, \ldots, 2(k+4)-1\}$ by placing the new elements $\{2 k+1,2 k+3,2 k+5,2 k+7\}$ into the sets such that $2 k+1$ and $2 k+7$ go into one set and $2 k+3$ and $2 k+5$ go into the other. This completes the induction. Now assume that $k$ is odd. Write $k=p m$. It is sufficient to find a subset $B$ of $\{1,3, \ldots, 2 k-1\}$ whose sum of elements is $p m^{2}$, because then the sum of the elements in the set $A=\{1,3, \ldots, 2 k-1\} \backslash B$ is $k^{2}-p m^{2}=p^{2} m^{2}-p m^{2}=(p-1) p m^{2}$, which is exactly $(p-1)$ times the sum of the elements in $B$. Consider the set $B=\{p, 3 p, \ldots,(2 m-1) p\} \subset\{1,3, \ldots, 2 k-1\}$. Then the sum of the elements in $B$ is

$$p+3 p+\cdots+(2 m-1) p=p(1+3+\cdots+(2 m-1))=p m^{2}$$

exactly as intended.