jmc

We can solve this problem using modular arithmetic. Donut Haven needs to fry at least $44\cdot 13$ donuts. Working modulo $20$, we have $$\begin{array}{rl}44\cdot 13& \equiv 4\cdot 13\\ & =52\\ & \equiv 12(\mathrm{mod}20).\end{array}$$Therefore, the number of donuts in $44$ full boxes is $12$ more than a multiple of $20$, which means it's $8$ less than the next multiple of $20$. Donut Haven will have $\boxed{8}$ donuts left over.