解：$(1)$设甲种货车每辆可装$x$吨，乙种货车每辆可装$y$吨.
根据题意，得$\{\begin{array}{l}2x+3y=13\\5x+6y=28.\end{array}，$
解方程组得$\{\begin{array}{l}x=2\\y=3.\end{array}.$
答：甲、乙两种货车每辆可分别装$2$吨、$3$吨.
$(2)50×(8×2+6×3)=1700($元).
答：货主应付货款$1700$元.
$(3)$设租用甲种货车共$a$辆，乙种货车$b$辆.
根据题意，得$2a+3b=20，$
此方程的非负整数解共有四个：
$\{\begin{array}{l}a=10\\b=0\end{array}.\{\begin{array}{l}a=7\\b=2\end{array}.\{\begin{array}{l}a=4\\b=4\end{array}.\{\begin{array}{l}a=1\\b=6.\end{array}.$
答：共有如下表所示的四种方案：
