The maximum sum of the series is:

Solution outline:

Interpret the series as an arithmetic progression where the first term is 20 and each term decreases by 2/3.

• General term: a_n = 20 + (n - 1)(-2/3).

• Find the largest n with a_n ≥ 0: 20 + (n - 1)(-2/3) ≥ 0 ⇒ (n - 1) ≤ 30 ⇒ n ≤ 31. So the last nonnegative term is the 31st term (which equals 0).

• Sum of first 31 terms: S_31 = (31/2)(a_1 + a_31) = (31/2)(20 + 0) = 31/2 × 20 = 310.

• Including any further terms (which are negative) would decrease the total, so the maximum possible sum is 310.

Answer: 310