Solution:
(I) Proof: According to the problem, we have $S_{n+1}-S_{n}= \frac {n+1}{3n}\cdot a_{n}$,
$\therefore a_{n+1}= \frac {n+1}{3n}\cdot a_{n}$,
$\therefore \frac {a_{n+1}}{n+1}= \frac {1}{3}\cdot \frac {a_{n}}{n}$,
Since $a_{1}=1$,
$\therefore$ the sequence $\left\{ \frac {a_{n}}{n}\right\}$ is a geometric sequence with the first term $1$ and common ratio $\frac {1}{3}$,
(II) From (I), we have $\frac {a_{n}}{n}=\left( \frac {1}{3}\right)^{n-1}$,
$\therefore a_{n}=n\cdot\left( \frac {1}{3}\right)^{n-1}$,
$\therefore S_{n}=1\times\left( \frac {1}{3}\right)^{0}+2\times\left( \frac {1}{3}\right)^{1}+3\times\left( \frac {1}{3}\right)^{2}+\ldots+n\cdot\left( \frac {1}{3}\right)^{n-1}$,
$\therefore \frac {1}{3}S_{n}=1\times\left( \frac {1}{3}\right)^{1}+2\times\left( \frac {1}{3}\right)^{2}+3\times\left( \frac {1}{3}\right)^{3}+\ldots+n\cdot\left( \frac {1}{3}\right)^{n}$,
$\therefore \frac {2}{3}S_{n}=1+\left( \frac {1}{3}\right)^{1}+\left( \frac {1}{3}\right)^{2}+\left( \frac {1}{3}\right)^{3}+\ldots+\left( \frac {1}{3}\right)^{n-1}-n\cdot\left( \frac {1}{3}\right)^{n}= \frac {1- \frac {1}{3^{n}}}{1- \frac {1}{3}}-n\cdot\left( \frac {1}{3}\right)^{n}= \frac {3}{2}-\left( \frac {3}{2}+n\right)\cdot\left( \frac {1}{3}\right)^{n}$,
$\therefore S_{n}= \frac {9}{4}-\left( \frac {9}{4}+ \frac {3n}{2}\right)\cdot\left( \frac {1}{3}\right)^{n}$．

Thus, the sum of the first $n$ terms of the sequence $\{a_{n}\}$, $S_{n}$, is $\boxed{\frac {9}{4}-\left( \frac {9}{4}+ \frac {3n}{2}\right)\cdot\left( \frac {1}{3}\right)^{n}}$.