1. **Problem statement:** We need to find all positive prime numbers $p$ such that $2p+1$ is a perfect cube. 2. **Formula and approach:** Let $2p+1 = n^3$ for some integer $n$. Since $p$ is prime and positive, we want to find all $p$ such that: $$2p + 1 = n^3$$ Rearranging gives: $$p = \frac{n^3 - 1}{2}$$ We need $p$ to be a positive prime number. 3. **Check values of $n$:** Since $p$ must be positive, $n^3 - 1 > 0 \Rightarrow n^3 > 1 \Rightarrow n > 1$. 4. **Test values of $n$ starting from 2:** - For $n=2$: $p = \frac{8 - 1}{2} = \frac{7}{2} = 3.5$ (not an integer, discard) - For $n=3$: $p = \frac{27 - 1}{2} = \frac{26}{2} = 13$ (13 is prime, keep) - For $n=4$: $p = \frac{64 - 1}{2} = \frac{63}{2} = 31.5$ (not integer, discard) - For $n=5$: $p = \frac{125 - 1}{2} = \frac{124}{2} = 62$ (not prime, discard) - For $n=6$: $p = \frac{216 - 1}{2} = \frac{215}{2} = 107.5$ (not integer, discard) - For $n=7$: $p = \frac{343 - 1}{2} = \frac{342}{2} = 171$ (not prime, discard) - For $n=8$: $p = \frac{512 - 1}{2} = \frac{511}{2} = 255.5$ (not integer, discard) - For $n=9$: $p = \frac{729 - 1}{2} = \frac{728}{2} = 364$ (not prime, discard) 5. **Check parity of $n^3 - 1$:** For $p$ to be integer, $n^3 - 1$ must be even. Since $n^3$ is odd if $n$ is odd, $n^3 - 1$ is even only if $n$ is odd. 6. **Check odd $n$ values:** We tested $n=3,5,7,9$ above. Only $n=3$ gives prime $p=13$. 7. **Conclusion:** The only positive prime $p$ such that $2p+1$ is a perfect cube is $p=13$. 8. **Sum of all such $p$:** Since there is only one, the sum is $13$. **Final answer:** $$\boxed{13}$$