Assuming that "several" means more than $1$, the only possible prime should be $p=3$. Let the number of blocks be $b$, where $1 < b < p$ and the common sum be $S$. Then $$bS = \sum_{i=1}^p n \equiv 0 \pmod p\implies p | S$$as $b < p$. Now the first block of numbers has to be $1, 2, ..., k$ where $k<p$ with the sum $\tfrac{k(k+1)}{2}$, divisible by p if and only if $p | (k + 1)\implies k = p-1$. Thus, the second block can only contain $p$, and we want $p = \tfrac{(p-1)p}{2}\implies p=3$. Construction for $p=3$ is just $\{1, 2\}, \{3\}$ NOTE: The choice of symbols was not a coincidence. :p