Find the sum.
\[
\sum_{n=1}^{60}\left(9-\frac{1}{3} n\right)
\]

The sum is $\square$.
(Type an integer or a decimal.)

Step 1 ：Given an arithmetic series with first term \(a_1 = 9 - \frac{1}{3} = 8.67\) (approx), common difference \(d = -\frac{1}{3}\), and number of terms \(n = 60\).

Step 2 ：The sum \(S\) of an arithmetic series is given by the formula: \(S = \frac{n}{2} * (a_1 + a_n)\), where \(a_n\) is the nth term of the series.

Step 3 ：We first need to find \(a_n\). In an arithmetic series, the nth term \(a_n\) is given by the formula: \(a_n = a_1 + (n - 1)*d\).

Step 4 ：Substituting the given values, we get: \(a_n = 8.67 - (60 - 1)*(-\frac{1}{3}) = 8.67 + \frac{59}{3} = 8.67 + 19.67 = 28.34\) (approx).

Step 5 ：Now, we can find the sum \(S\): \(S = \frac{60}{2} * (8.67 + 28.34) = 30 * 37.01 = 1110.3\) (approx).

Step 6 ：So, the sum of the series is approximately \(\boxed{1110.3}\).