 Solution

The ground state changes from high-spin $t_{2g}^3{e}_g^2$, ${}^6{A}_{1g}$, to low-spin $t_{2g}^5$, ${}^2{T}_{2g}$. The larger splitting eventually compensates the additional pairing cost. Because all plotted excitation energies are re-referenced to the new ground state, the branches change slope at the crossover. The break is not an infinite physical excitation energy or a discontinuity of the underlying Hamiltonian.