In the case of diamagnetism, the presence of an externally applied magnetic field $\mathbf{B}$ will induce a magnetisation $\mathbf{M}$ that acts opposite to the direction of the field. This is why diamagnetic materials expel magnetic flux from the bulk of the material. Therefore, in order to understand why the susceptibility ($\chi$) is negative, it is important to know why the magnetic moments of the atoms ($\mu$) anti-align with the field (thus resulting in $\mathbf{M}$ antiparallel to $\mathbf{B}$).
Semi-classical intuition: Lenz's law
(Disclaimer: Whilst this provides a rough semi-classical intuition of why $\chi < 0$. The true explanation requires quantum mechanics, which I have alluded to at the end) Consider an electron in a circular orbit around the nucleus at radius $r$. Now we switch on a magnetic field, so that the field rises from 0 to $\mathbf{B}$ in time $\delta t$. Using Lenz's law, we can determine the electric field acting on the electron due to the change in magnetic flux:
\begin{equation}
\oint_{circle} \mathbf{E}.d\mathbf{l} = -\frac{\partial \Phi_{B}}{\partial t}
\end{equation}
With magnetic flux $\Phi_{B} = B\pi r^2$ :
\begin{equation}
E = -\frac{Br}{2\delta t}
\end{equation}
This electric field will exert a torque on the electron which, as you can check, increases the angular momentum by $\delta L$, where:
\begin{equation}
\delta L = \frac{eBr^2}{2}
\end{equation}
The electron is travelling in a circular current loop, so this increase in angular momentum will change the magnetic dipole moment by $\delta \mu$. From magnetostatics, we have $\delta \mu = I (\pi r^2)$, where $I$ is the electric current. Since current is the rate of flow of charge:
\begin{equation}
I = \frac{-e}{T}
\end{equation}
where $T$ is the time-period for one orbit.
\begin{equation}
T = \frac{2\pi r}{v} = \frac{2\pi m r^2}{\delta L}
\end{equation}
Plugging everything into the expression for $\delta \mu$, we find:
\begin{equation}
\delta \mu = -\frac{e^2 B r^2}{4 m}
\end{equation}
Therefore, the magnetisation $M$ (total magnetic moment per unit vol.) is given by:
\begin{equation}
M = -\frac{\rho e^2 B r^2}{4m}
\end{equation}
where $\rho$ is the number of atoms per unit vol. The negative sign is crucial, because when you take the derivative with respect to $H (= B / \mu_{0})$, you get the susceptibility $\chi$:
\begin{equation}
\chi = -\frac{\mu_{0} \rho e^2 r^2}{4m}
\end{equation}
With Quantum Mechanics
Of course, electrons don't travel in circular orbits around the nucleus. Instead, they exist in orbitals/ wavefunctions around the nucleus. This means that we can only meaningfully speak of $\langle r^2 \rangle$ in the above expression. Assuming the magnetic field is aligned in the $z$-direction, the electron will be moving in the $xy$-plane, so we need $\langle x^2 + y^2 \rangle$. Assuming the atom is spherically symmetric:
\begin{equation}
\langle x^2 + y^2 \rangle = \frac{2}{3} \langle r^2 \rangle
\end{equation}
Using this, instead of $r^2$ in the expression for $\chi$, we get:
\begin{equation}
\chi = - \frac{\mu_{0} \rho e^2 \langle r^2 \rangle}{6m}
\end{equation}
A proper treatment of the quantum mechanics involves deriving the Hamiltonian for an electron in a magnetic field, before applying 1st order perturbation theory to the diamagnetic term.
