= Solution
\b[Perfect identification on every trial is impossible], because the family contains a nonorthogonal pair. Any physical identification device induces a <POVM> whose effect $E_i$ corresponds to its declaration of label $i$. Certainty of correct identification would require $\langle\psi_i|E_i|\psi_i\rangle=1$ and $\langle\psi_j|E_i|\psi_j\rangle=0$ for every $j\ne i$.
Since $0\leq E_i\leq I$, the first identity gives $(I-E_i)|\psi_i\rangle=0$, while positivity and the second identity give $E_i|\psi_j\rangle=0$. For example these conclusions follow by writing the zero expectations as squared norms of $(I-E_i)^{1/2}\psi_i$ and $E_i^{1/2}\psi_j$. Thus
$$
\langle\psi_j|\psi_i\rangle=\langle\psi_j|E_i|\psi_i\rangle=0.
$$
Perfect identification would force the entire family to be <orthogonal>, contrary to the hypothesis. This argument permits ancillas, general measurements and disturbance of the input; it does not assume that the device must preserve the state.
\b[Unambiguous identification with a nonzero success <probability> for every input is possible.] Normalize the given kets and let $G_{ij}=\langle\psi_i|\psi_j\rangle$ be their <Gram matrix>. Their <linear independence> makes $G$ positive definite. Define reciprocal vectors within their span by
$$
|\chi_i\rangle=\sum_j(G^{-1})_{ji}|\psi_j\rangle.
$$
Then $\langle\psi_k|\chi_i\rangle=\sum_jG_{kj}(G^{-1})_{ji}=\delta_{ki}$, and hence also $\langle\chi_i|\psi_k\rangle=\delta_{ik}$. Put $R=\sum_i|\chi_i\rangle\langle\chi_i|$. It is positive and has finite <operator norm>. Choosing $\epsilon=1/(2\|R\|)>0$, set
$$
\boxed{E_i=\epsilon|\chi_i\rangle\langle\chi_i|,\qquad E_?=I-\epsilon R.}
$$
All these effects are positive, and $E_?\geq I/2$, so they sum to the identity and define a <POVM>. The <Born rule> gives
$$
\boxed{\mathbb P(i\mid\psi_k)=\epsilon\delta_{ik},\qquad
\mathbb P(?\mid\psi_k)=1-\epsilon.}
$$
Thus every label is correct whenever it occurs and every input has the strictly positive success <probability> $p_k=\epsilon$. This is the <reciprocal-state construction of unambiguous discrimination>.
For an explicit success constant, collect the input kets into a matrix $\Psi$, so $\Psi^\dagger\Psi=G$. Its polar factorization is $\Psi=UG^{1/2}$ with $U^\dagger U=I_N$. The reciprocal columns are $\Psi G^{-1}=UG^{-1/2}$, and therefore $R=UG^{-1}U^\dagger$. Its largest <eigenvalue> is $1/\lambda_{\min}(G)$, so one can take $\epsilon=\lambda_{\min}(G)/2$. Since the normalized <Gram matrix> has trace $N$, this is at most $1/2$.
To realize the device, use measurement operators $M_i=\sqrt{E_i}$ and $M_?=\sqrt{E_?}$. The map $|v\rangle\mapsto\sum_iM_i|v\rangle|i\rangle+M_?|v\rangle|?\rangle$ preserves <inner products> because the effects sum to $I$. Extend this <linear isometry> to a unitary on the input and an initially blank ancilla, then projectively measure the ancilla's <orthogonal> outcome labels. Its outcome <probabilities> are precisely those above. This gives a physical <unambiguous quantum state discrimination> device rather than only a formal set of effects.
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