Let $(b_i)_{1\leq i\leq p+1}$ be a permutation of $(a_i)_{1\leq i\leq p+1}$ such that $\sum_{i=1}^{p}b_i$ and $\sum_{i=1}^{p-1}ib_i$ aren't divisible by $p$. Denote $M=\sum_{i=1}^{p-1}ib_i$ and $X=\sum_{i=1}^{p}b_i$. So it is clear that: $M+X= b_p +2b_1+\cdots+(p-1)b_{p-2} +pb_{p-1}\equiv b_p +2b_1+\cdots+(p-1)b_{p-2}\: (mod\: p)$; $M+2X= 2b_p +3b_1+\cdots+pb_{p-2} +(p+1)b_{p-1}\equiv b_{p-1} +2b_p+3b_1+\cdots+(p-1)b_{p-3}\: (mod\: p)$; And so on. Hence: $M+kX\equiv b_{p-k} +2b_{p-k+1}+\cdots+ kb_p+(1+k)b_{1}+\cdots+(p-1)b_{p-k-1}\: (mod\: p)$, As we are suposing that $X\not\equiv 0\:(mod\: p)$ it is true that there is some $k$, ($1\leq k \leq p-1$), such that $M+kX\equiv 0\:(mod\: p)$ and then: $b_{p-k} +2b_{p-k+1}+\cdots+ kb_p+(1+k)b_{1}+\cdots+(p-1)b_{p-k-1}\equiv 0\:(mod\: p$) so we are done.