Answer:
Solution:
The greatest of Winnie's even numbers is $2\times 30 = 60,$ so her sum is
\[2+4+6+\cdots + 58 + 60.\]For Grogg's sum, there are five odd numbers between $0$ and $-10$, five between $-10$ and $-20$, and so on, so Grogg's sum is
\[(-1)+(-3)+(-5)+\cdots+(-57)+(-59).\]
We could compute these sums separately and find their difference, but instead, we'll look for a clever way to regroup to compute the difference
\[(2+4+6+\cdots +58+60)-((-1)+(-3)+(-5)+\cdots +(-57)+(-59)).\]Noting that Winnie's list steps up by 2 at each term while Grogg's steps down by 2 at each step, we write Winnie's list backwards, so that each list goes down by 2 at each step:
\[(60+58+56+\cdots + 4+2) - ((-1)+(-3)+(-5)+\cdots +(-57)+(-59)).\]Now, we see that the difference between a term in Winnie's backwards list and the corresponding term in Grogg's list is always the same:
\begin{align*}
60-(-1) &= 61,\\
58-(-3) &= 61,\\
56-(-5) &=61,\\
\vdots&\\
2-(-59) &= 61.
\end{align*}There are 30 such differences, so Winnie's sum is $30\times 61 = \boxed{1{,}830}$ greater than Grogg's sum.
Explanation: