Lets denote $(p_n)$ the sequence of all prime numbers $(p_1=2, p_2=3,\ldots)$.
The conjecture is the following.
For $n$ odd and greater than $2$,
$$\exists I \subset \{1,\ldots ,n\} \quad \sum_{i\in I } p_i=\sum_{i\in\{1,\ldots,n\}\setminus I} p_i.$$
For instance, $n=3$ works because $2+3=5$ $(I=\{1,2\})$.
But also $n=5$ because $2+5+7=3+11$ $(I=\{1,2\})$.
And also $n=7$ because $2+3+7+17=5+11+13$ $(I=\{1,2\})$.
We can prove that it can't work for $n$ even by parity (we would have an odd number of odd prime numbers).
We can reformulate the problem like this:
For $n$ odd and greater than $2$,
$$\exists I \subset \{1,\ldots ,n\} \quad \sum_{i\in I } p_i=\frac 12\sum_{i\leqslant n} p_i.$$
I honestly don't know how to start here, and I don't even know if this could be true. Do you have any leads ?