Answer: $-18910$

Change the order of summation and simplify the inner sum:
$$
\begin{aligned}
\sum_{k=1}^{60} \sum_{n=1}^{k} \frac{n^{2}}{61-2 n} & =\sum_{n=1}^{60} \sum_{k=n}^{60} \frac{n^{2}}{61-2 n} \\
& =\sum_{n=1}^{60} \frac{n^{2}(61-n)}{61-2 n}
\end{aligned}
$$

Then, we rearrange the sum to add the terms corresponding to $n$ and $61-n$ :
$$
\begin{aligned}
\sum_{n=1}^{60} \frac{n^{2}(61-n)}{61-2 n} & =\sum_{n=1}^{30}\left(\frac{n^{2}(61-n)}{61-2 n}+\frac{(61-n)^{2}(61-(61-n))}{61-2(61-n)}\right) \\
& =\sum_{n=1}^{30} \frac{n^{2}(61-n)-n(61-n)^{2}}{61-2 n} \\
& =\sum_{n=1}^{30} \frac{n(61-n)(n-(61-n))}{61-2 n} \\
& =\sum_{n=1}^{30}-n(61-n) \\
& =\sum_{n=1}^{30} n^{2}-61 n
\end{aligned}
$$

Finally, using the formulas for the sum of the first $k$ squares and sum of the first $k$ positive integers, we conclude that this last sum is
$$
\frac{30(31)(61)}{6}-61 \frac{30(31)}{2}=-18910
$$

So, the original sum evaluates to -18910 .
11. $[7]$ FInd the value of
$$
\sum_{k=1}^{60} \sum_{n=1}^{k} \frac{n^{2}}{61-2 n}
$$