$v_p(i^p+(p-i)^p)=2$ so : $p^2|i^p+(p-i)^p \implies r_i+r_{p-i}=p^2$ which implies the result.

notice that $r_i+r_{p-i}$ divisible by $p^2$ due to lifting exponent since $\gcd(r_i,p)=1$ and $v_p(i^p+(p-i)^p) = v_p(p)+v_p(p) = 2$ now we know that since it is a remainder of $p^2$ and there is no equal to zero $0 <r_i <p^2$ $0<r_{p-i}<p^2$ $0<r_i+r_{p-i}<2p^2$ but $r_i+r_{p-i}$ multiple of $p^2$ automatically this force $r_i+r_{p-i}= p^2$ then sum all of this term $r_1+r_{p-1} = p^2$ . . . . . $r_{p-1}+r_1 = p^2$ $2r_1+2r_2+...+2r_{p-1} = p^2(p-1)$ $r_1+.....+r_{p-1} = \frac{p^2(p-1)}{2}$