I don't think that's possible: Suppose your initial state is $\vert \phi\rangle$. Let $\vert \psi\rangle$ be a state with the numbers $1,4,8,13$ in superposition. Then $\langle \psi \vert \phi\rangle\neq 0$, since the states have $8$ and $13$ in common. But if we have some sort of $n$th minimum circuit $U_n$ with $U_2\vert \phi\rangle\vert 0\rangle=\vert\phi\rangle\vert 3\rangle$ and $U_2\vert \psi\rangle\vert 0\rangle=\vert \psi\rangle\vert 4\rangle$, the outputs are orthogonal. This is impossible if $U_2$ is a unitary.
I'm not sure exactly what you had in mind (a circuit, a channel,...?). In the most generality I can think of: Suppose $\vert \phi\rangle$ and $\vert\psi\rangle$ are uniform superpositions of the numbers $\{1,3,8,13\}$ and $\{1,4,8,13\}$, respectively. Then the fidelity between them is 0.75. Suppose we have some quantum channel $\Phi$ that is supposed to select the 2nd smallest element. We know the fidelity between $\Phi(\vert \phi\rangle\langle\phi\vert)$ and $\Phi(\vert \psi\rangle\langle\psi\vert)$ is at least 0.75 because channels increase fidelity. (I think) This implies that if the probability of measuring 3 in the first state is $p$, then the probability of measuring 3 in the second state is at least $0.75p$. So if we have a high probability of measuring 3 in the first state (which we should) then we also have a high probability of measuring 3 in the second state (which we should not).
Note that there is a big difference between what you asked, and the problem of finding the $n$th minimum given the state $\vert a_1\rangle\vert a_2\rangle\cdots \vert a_N\rangle$. In this case, if any of the numbers $a_1,\cdots, a_N$ are different, the states are orthogonal. The paper you linked provides a method to solve the problem in this case.
