The maximum sum of the series $20+19\frac{1}{3}+18\frac{2}{3}+\cdots$ is

nth term of the series is 20 + (n - 1) (-2/3).
For the sum to be maximum, nth term $\ge$ 0

$\begin{array}{l}\Rightarrow 20+(\mathrm{n}-1)\left(-\frac{2}{3}\right)\ge 0\\ \Rightarrow \mathrm{n}\le 31\end{array}$

Thus, the sum of 31 terms is maximum and is $\frac{31}{2}\left[40+30\times \left(-\frac{2}{3}\right)\right]=310$