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

\(S_N=\dfrac{1}{3}\left(\dfrac14-\dfrac{1}{3N+4}\right)\), so \(S=\dfrac{1}{12}\).

Therefore \(S-S_N=\dfrac{1}{3(3N+4)}\).

We need \(\dfrac{1}{3(3N+4)}\lt\dfrac{1}{10000}\), giving \(N\gt1109\dfrac{7}{9}\). Hence the least integer value is \(N=1110\).

Start with the right hand side:

\(\frac{1}{3}\left(\frac{1}{3r+1}-\frac{1}{3r+4}\right)=\frac{1}{3}\left(\frac{(3r+4)-(3r+1)}{(3r+1)(3r+4)}\right)=\frac{1}{(3r+1)(3r+4)}.\)

Therefore

\(S_N=\sum_{r=1}^{N}\frac{1}{(3r+1)(3r+4)}=\frac{1}{3}\sum_{r=1}^{N}\left(\frac{1}{3r+1}-\frac{1}{3r+4}\right).\)

Writing out the terms gives a telescoping sum:

\(S_N=\frac{1}{3}\left[\left(\frac14-\frac17\right)+\left(\frac17-\frac1{10}\right)+\cdots+\left(\frac{1}{3N+1}-\frac{1}{3N+4}\right)\right].\)

All the middle terms cancel, so

\(S_N=\frac{1}{3}\left(\frac14-\frac{1}{3N+4}\right).\)

Taking the limit as \(N\to\infty\),

\(S=\frac{1}{3}\cdot\frac14=\frac{1}{12}.\)

Hence

\(S-S_N=\frac{1}{12}-\frac{1}{3}\left(\frac14-\frac{1}{3N+4}\right)=\frac{1}{3(3N+4)}.\)

We require

\(\frac{1}{3(3N+4)}\lt\frac{1}{10000}.\)

This gives

\(3(3N+4)\gt10000,\)

so

\(9N+12\gt10000,\qquad N\gt\frac{9988}{9}=1109\frac{7}{9}.\)

Therefore the least integer value of \(N\) is \(1110\).