The statistical definition of temperature is the following:
$$\frac{1}{k_BT} = \frac{d\ln(\Omega)}{dE}$$
where $k_B$ is the Boltzmann constant, $T$ is the temperature, $\Omega$ is the possible number of states, and $E$ is the energy. Rearranging for T we have the following:
$$\begin{split} T & = \frac{1}{k_B} \frac{dE}{d\ln(\Omega)} \\[0.5em] T & \propto \frac{dE}{d\ln(\Omega)} \\[0.5em] \end{split}$$
The differential on the right handside: $\frac{dE}{d\ln(\Omega)}$ implies that if there was a negative change in energy with an increase in microstates, or a decrease in energy with an increase in microstates, the temperature of the system would be negative.
Can this occur, and if so what does it mean physically for the system?