Verify that \(\dfrac{1}{(3r+1)(3r+4)}=\dfrac{1}{3}\left(\dfrac{1}{3r+1}-\dfrac{1}{3r+4}\right)\).

Let \(S_N\) denote \(\sum_{r=1}^{N}\dfrac{1}{(3r+1)(3r+4)}\) and let \(S\) denote \(\sum_{r=1}^{\infty}\dfrac{1}{(3r+1)(3r+4)}\). Find the least value of \(N\) such that \(S-S_N\lt\dfrac{1}{10000}\).