Orbitals and Pauli Exclusion Principle without Slater determinants
I think what you are asking is "What form does $\Psi$ need to take so that the exchange interaction exists when the electronic Hamiltonian given acts upon it?". (Noting that the exchange interaction is due to the fact that electrons are indistinguishable and antisymmetric with respect to their positions). 

From what I understand the Hamiltonian you have given acts upon electrons and so does not include the exchange interactions. As far as I'm aware the analytical form of the electronic Hamiltonian that includes exchange is not known. If it were, then density functional theory would not have so many approximate exchange-correlation functionals.

I will try to show the problem by presenting the form of the Schrodinger Equation in a slightly different way than is presented in Szabo and Ostlund's book. 

Firstly, it is understood that $N$ in the Hamiltonian you give in the question represents electrons - not spin orbitals and we know that the electron density $\rho$ at $\textbf{r}$ is represented: \begin{equation} \label{eq1} \rho(\textbf{r}) = |\Psi(\textbf{r})|^2 = \langle \Psi|\textbf{r}\rangle\langle \textbf{r}|\Psi \rangle = \Psi^*(\textbf{r})\Psi(\textbf{r}) \end{equation}
The Hamiltonian you give acts upon the electron density by using spin orbitals to represent it.

Note, a physical electron density must therefore arise from a wavefunction that is anti-symmetric. 

The Hamiltonian you have given can be seen to act upon the wavefunction as follows:
\begin{align}
E = &amp;\int \Psi^*(\textbf{r}_1)(-\frac{1}{2}\nabla^2-\sum_{A=1}^{M}\frac{Z_A}{|\textbf{r}_1-\textbf{r}_A|}) \Psi(\textbf{r}_1) d\textbf{r}_1 \\ &amp;+\frac{1}{2}\int\int \Psi^*(\textbf{r}_1) \Psi^*(\textbf{r}_2) (\frac{1}{|\textbf{r}_1-\textbf{r}_2|}) \Psi(\textbf{r}_1)\Psi(\textbf{r}_2)d\textbf{r}_1d\textbf{r}_2
\end{align}
Note that the second term - concerning two-electron integrals shows the electron density at $\textbf{r}_1$ interacting with the electron density of $\textbf{r}_2$; this is clear when rearranging the final term (knowing that the $r^{-1}_{12}$ operator is a Hermitian operator) and substituting in the first equation: 
\begin{align}
\langle 12|12\rangle &amp;=\int\int \Psi^*(\textbf{r}_1) \Psi(\textbf{r}_1) (\frac{1}{|\textbf{r}_1-\textbf{r}_2|}) \Psi^*(\textbf{r}_2)\Psi(\textbf{r}_2)d\textbf{r}_1d\textbf{r}_2 \\ &amp;= \int\int \rho(\textbf{r}_1) (\frac{1}{|\textbf{r}_1-\textbf{r}_2|}) \rho(\textbf{r}_2)d\textbf{r}_1d\textbf{r}_2 
\end{align}
Now, it is well known that electrons are indistinguishable, (i.e. they exist with a given probability density over all space and an electron cannot be labelled) such that: $$ |\Psi(1,2)|^2 = |\Psi(2,1)|^2 $$ Where $1=\textbf{r}_1$ and $2=\textbf{r}_2$.

It is also known that, as fermions (half-integal spin particles) they are anti-symmetric:
 $$ \Psi(1,2) = -\Psi(2,1) $$ 
Applying this antisymmetry to the two electron integrals we know that there has to be an antisymmetric electron interaction due to the interchanging of electronic positions. Therefore, the two electron integrals can be written: $$ \langle 12||12\rangle = \langle 12|12 \rangle -\langle 12|21 \rangle $$ $$ \langle 21||21\rangle = \langle 21|21 \rangle -\langle 21|12 \rangle $$ both the negative terms are the exchange interactions associated with the interchanging positions. Which are not included in the electronic Hamiltonian given in the question and their integral forms cannot be rearranged as shown before into terms of the electron density at a single given position.

This is why the Fock Operator is used instead of the electronic Hamiltonian!

I hope this all makes sense! I have probably made a mistake somewhere! So please let me know! 



If you would like to know more about calculating the exchange from the electron density for known wavefunctions I recommend the paper: Pendás, A. M.; Blanco, M. A.; Francisco, E. Two-electron integrations in the quantum theory of atoms in molecules. J. Chem. Phys. 2004, 120 (10), 4581–4592. DOI: 10.1063/1.1645788.

For normailised wavefunctions, the exchange is given:
\begin{equation}
K = - \int \int r_{12}^{-1} \vert \rho_{1}(\textbf{r}_{1};\textbf{r}_{2}) \vert^{2}d\textbf{r}_{1}d\textbf{r}_{2}
\end{equation}

Where:
$$\rho_{1}(\textbf{r}_{1};\textbf{r}_{1}^{\prime}) = N_{\text{elec}} \int \Psi_{\text{elec}} (\textbf{r}_{1},...,\textbf{r}_{N_{\text{elec}}})\Psi_{\text{elec}}^{*}(\textbf{r}_{1}^{\prime},...,\textbf{r}_{N_{\text{elec}}}) d\textbf{r}_{2}...d\textbf{r}_{N_{\text{elec}}}$$