The "canonical" example of a map that is $k$-positive but not $(k+1)$-positive is the map defined by
$$
\Phi_k(X) = k\cdot\mathrm{Tr}(X)I_n - X.
$$

Above, $n$ denotes the size of $X$ (i.e., $X \in M_n$) and $I_n$ is the $n \times n$ identity matrix. This map was introduced in "J. Tomiyama. On the geometry of positive maps in matrix algebras II. Linear Algebra Appl., 69:169–177, 1985." (though it had been considered by Choi earlier in the special case when $k = n-1$).

It's not difficult to prove that this map is $k$-positive just by elementary linear algebra. I won't post details here since it's a bit long and messy, but it's almost exactly the same as the proof that starts at the bottom of page 4 of [these notes](http://www.njohnston.ca/wp-content/uploads/2008/12/PEB_2009.pdf).

To show that this map is <em>not</em> $(k+1)$-positive (when $k < n$), simply let $\mathbf{v} = \sum_{i=1}^{k+1} \mathbf{e}_i \otimes \mathbf{e}_i \in \mathbb{C}^{k+1} \otimes \mathbb{C}^n$ (here $\{\mathbf{e}_i\}$ is the standard basis of $\mathbb{C}^{k+1}$ on the first subsystem, and it's just embedded into $\mathbb{C}^n$ in the natural way on the second subsystem). Then compute that
$$
(id_{k+1} \otimes \Phi_k)(\mathbf{v}\mathbf{v}^*) = k(I_{k+1} \otimes I_n) - \mathbf{v}\mathbf{v}^*,
$$
which has $-1$ as an eigenvalue and is thus not positive.

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A couple of side notes:

 1. This map shows that Choi's theorem on complete positivity is optimal in some sense: if $k \geq n$ then $k$-positivity implies complete positivity, but if $k < n$ then $k$-positivity and $(k+1)$-positivity are indeed different sets.

 2. In the $k = 1$ case this map comes up frequently in quantum information theory under the name of the "reduction map".