Quoting from the linked source: "thus SWAP has negative eigenvalues, which means that $T\otimes I$ is not positive and therefore $T$ is not completely positive", where $T$ is the transpose. So they are not saying that the SWAP is not a realisable operation; they are saying that $T$ isn't.

As you note, the SWAP is a perfectly fine unitary gate. That there are negative eigenvalues is not a problem for a unitary operation (which has in general complex eigenvalues in the unit circle).

The argument made in the linked textbook is that if a quantum map $\Phi$ is such that $\Phi\otimes I$ is not positive, i.e. has some negative eigenvalue, then $\Phi$ is not completely positive, and thus does not describe a physical quantum operation.

Quoting from the linked source: "thus SWAP has negative eigenvalues, which means that $T\otimes I$ is not positive and therefore $T$ is not completely positive", where $T$ is the transpose. So they are not saying that the SWAP is not a realisable operation; they are saying that $T$ isn't.

As you note, the SWAP is a perfectly fine unitary gate. That there are negative eigenvalues is not a problem for a unitary operation (which has in general complex eigenvalues in the unit circle).

The argument made in the linked textbook is that if a quantum map $\Phi$ is such that $\Phi\otimes I$ is not positive, i.e. sends some positive semidefinite operator to some non-positive-semidefinite operator, then $\Phi$ is not completely positive, and thus does not describe a physical quantum operation.

Just to try to clarify a few things:

1. Let $V$ be a (complex) finite-dimensional vector spaces. Let $U\in\mathrm{Lin}(V)$ be a linear operator. We say that $U$ is positive semidefinite if it is Hermitian and has non-negative eigenvalues, i.e. $U^\dagger=U$ and $\langle v,Uv\rangle\ge0$ for all $v\in V$.

2. Let $\Phi$ be a (quantum) map on $V$. This is, by definition, a linear operator acting on linear operators, that is, $\Phi\in\mathrm{Lin}(\mathrm{Lin}(V),\mathrm{Lin}(V))$. You can think of it as an object sending density matrices (which are linear operators in $V$) to other density matrices.

3. A map $\Phi$ is said to be positive if it sends positive semidefinite operators to positive semidefinite operators, that is, $\Phi(X)\ge0$ whenever $\mathrm{Lin}(V)\ni X\ge0$. It is said to be completely positive if $\Phi\otimes I$ is positive for any possible extension.

4. So, they want to show that the transpose map $T$ is positive but not completely positive. This is the map sending $|i\rangle\!\langle j|$ to $|j\rangle\!\langle i|$ for all $i,j$. Note that here $|i\rangle\!\langle j|\in\mathrm{Lin}(V)$. To show that this is the case, they argue that, if $|\Psi\rangle\in V\otimes V$ denotes the maximally entangled state (note the enlarged space), then $(T\otimes I_V)\in\mathrm{Lin}(\mathrm{Lin}(V\otimes V),\mathrm{Lin}(V\otimes V))$ is a map sending $|\Psi\rangle\!\langle\Psi|$ (note that we use the density matrix corresponding to the ket state here) to the SWAP operator. Again, here $\text{SWAP}\in\mathrm{Lin}(V\otimes V)$. Now, this SWAP is not positive semidefinite (again, as a matrix/operator), hence $T\otimes I$ is not positive as a map, hence $T$ is not completely positive (again, as a map).