For a closed organ pipe, the relationship between the length \( L \) of the pipe and the harmonic number \( n \) is given by:
\[
L = \frac{(2n-1)\lambda}{4},
\]
where \( \lambda \) is the wavelength of the sound.
For the third overtone, we have:
\[
n = 4.
\]
This corresponds to the 4th harmonic, which has 4 nodes and 4 antinodes, with the pattern of oscillation as follows:
- The pipe has four full loops, plus a half loop.
- This results in 4 nodes and 4 antinodes.
Final Answer:
\[
\boxed{\text{Four nodes and four antinodes}}.
\]