jmc

To start, notice that $24\cdot 50=1200\equiv 1(\mathrm{mod}1199)$ (in other words, $24$ and $50$ are inverses modulo $1199$).

To solve the congruence $24x\equiv 15(\mathrm{mod}1199)$, we multiply both sides by $50$ and simplify: $$\begin{array}{rl}50\cdot 24x& \equiv 50\cdot 15(\mathrm{mod}1199)\\ x& \equiv 750(\mathrm{mod}1199)\end{array}$$This process can also be reversed (by multiplying both sides by $50^{-1}=24$), so the solutions to the original congruence are precisely the same as the solutions to $x\equiv 750(\mathrm{mod}1199)$. The largest negative solution is $750-1199=\boxed{-449}$.