If $n$ is odd, then$$\sum_{i=1}^{(n+1)/2} i (n-i) = \frac{(n-1)(n+1)(n+3)}{12}$$Hence, you get that $$\frac{S_n}{n^2} = \frac{2}{n^2(n+1)} \times \frac{(n-1)(n+1)(n+3)}{12} = \frac{(1-1/n)(1+3/n)}{6}$$
If $n$ is even, then$$\sum_{i=1}^{n/2} i (n-i) = \frac{n(n+2)(2n-1)}{24}$$Hence, you get that $$\frac{S_n}{n^2} = \frac{2}{n^2(n+1)} \times\frac{n(n+2)(2n-1)}{24} = \frac{\left(1+\frac1{n+1} \right) \left(1-\frac1{2n} \right)}{6}$$
