= Solution
The linear extension of the stated <k-reduction map> is
$$
\Lambda_k(X)=k\operatorname{Tr}(X)I-X.
$$
It suffices to consider a pure state $|v\rangle$ of <Schmidt rank> $r\leq k$, because positivity is preserved by sums. Write
$$
|v\rangle=\sum_{j=1}^r\sqrt{\lambda_j}|j\rangle|j\rangle,
\qquad
\rho_A=\sum_{j=1}^r\lambda_j|j\rangle\langle j|.
$$
Then
$$
(\operatorname{id}\otimes\Lambda_k)(|v\rangle\langle v|)
=k\rho_A\otimes I-|v\rangle\langle v|.
$$
For every $|x\rangle=\sum_{ij}x_{ij}|ij\rangle$, the <Cauchy-Schwarz inequality> gives
$$
|\langle v|x\rangle|^2
=\left|\sum_{j=1}^r\sqrt{\lambda_j}x_{jj}\right|^2
\leq r\sum_{j=1}^r\lambda_j|x_{jj}|^2
\leq k\langle x|\rho_A\otimes I|x\rangle.
$$
Thus the operator is a <positive semidefinite operator>. Applying this to every vector in a Schmidt-number-$k$ ensemble proves
$$
\boxed{\operatorname{SN}(\sigma)\leq k
\ \Longrightarrow\
(\operatorname{id}_n\otimes\Lambda_k)(\sigma)\geq0}.
$$
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