Regrouping makes it easier

Sum of { [2 × 60 – (2k – 1)] × k }
= Sum of { [121 – 2k] × k }
= Sum of { 121k – 2k² }

121 – 2k > 0, k < 60.5
k = [ 1 , 60 ]

Sum of { 121k – 2k² } for 1 ≤ k ≤ 60
= 121 × [k(k+1)/2] – 2 × [k(k+1)(2k+1)/6]
= k(k + 1) × [ 121/2 – (2k + 1)/3 ]
= 60 × 61 × [ 121/2 – 121/3 ]
= 73810

When $n=60$ , the sum of positive terms becomes:

$\begin{aligned} S & = 119 \cdot 1 + 117 \cdot 2 + 115 \cdot 3 + \cdots + 1 \cdot 60 \\ & = \sum_{k=1}^{60} (121-2k)k \\ & = 121 \sum_{k=1}^{60} k - 2 \sum_{k=1}^{60} k^2 \\ & = 121 \cdot \frac {60\cdot 61}2 - 2 \cdot \frac {60 \cdot 61 \cdot 121}6 \\ & = 60 \cdot 61 \cdot 121 \left( \frac 12 - \frac 13 \right) \\ & = 10 \cdot 61 \cdot 121 \\ & = \boxed{73810} \end{aligned}$

$\begin{array}{l}The\ series\ can\ be\ written\ as:\\ 2n-1\\ 2n-3\ \ \ +\ \ \ 2n-3\ \ \\ 2n-5\ \ \ +\ \ \ 2n-5\ \ \ +\ \ \ 2n-5\\ ...\\ ...\\ ...\\ \ \ \ \overbrace{5\ \ \ \ \ \ \ \ \ +\ \ \ \ \ \ \ \ 5\ \ \ \ \ \ \ +\ \ \ \ \ \ 5\ \ \ \ \ \ \ \ +......+\ \ \ \ 5}^\text{n-2 times}\ \ \ \ \ \ \\ \ \ \\\overbrace{\ \ \ 3\ \ \ \ \ \ \ \ \ +\ \ \ \ \ \ \ 3\ \ \ \ \ \ \ \ +\ \ \ \ \ \ 3\ \ \ \ \ \ \ \ +......+\ \ \ \ 3\ \ \ \ \ +\ \ \ \ \ 3}^\text{n-1 times}\\ \ \ \ \overbrace{1\ \ \ \ \ \ \ \ \ +\ \ \ \ \ \ 1\ \ \ \ \ \ \ \ \ +\ \ \ \ \ \ 1\ \ \ \ \ \ \ \ +......+\ \ \ \ 1\ \ \ \ \ +\ \ \ \ \ 1\hspace{0.5cm}+\hspace{0.5cm}1 }^\text{n times}\\\rule{13cm}{0.01cm}\\ \ \ \ n^2\ \ \ \ \ \ \ \ +\ \ (n-1)^2\ +\ (n-2)^2 +......+\ \ \ \ 3^2\ \ \ +\ \ \ \ \ 2^2\ \ \ +\ \ \ \ 1^2\\Adding\ \ columnwise,\ \ we\ \ get\ \ the\ \ sum\ \ of\ \ consecutive\ \ odd\ \ numbers\ \ in\ \ each\ \ column.\\As\ \ the\ \ sum\ \ of\ \ the\ \ first\ \ k\ \ odd\ \ numbers\ \ is\ \ k^2,\ \ the\ \ problem\ \ reduces\ \ to\ \ finding\ \ the\ \ sum\ \ of\ \ the \ \ first\ \ n\ \ squares.\\Hence,\ \ the\ \ sum\ \ is\ \ (n(n+1)(2n+1))/6.\\Plugging\ \ in\ \ n=60,\ \ we\ \ get\ \ 73810. \end{array}$