Saudi Arabian Mathematical Competitions

For $j=1,2, \ldots, p-1$, we have
$$\frac{j^{p-1}-1}{p}=a_{j} p+r_{j}$$
for some integer $a_{j}$. It follows
$$\frac{j^{p}-j}{p}=j a_{j} p+j r_{j},$$
hence
$$\frac{j^{p}-j+(p-j)^{p}-(p-j)}{p}=j a_{j} p+j r_{j}+(p-j) a_{p-j} p+(p-j) r_{p-j}.$$
We obtain
$$\frac{j^{p}+(p-j)^{p}}{p}=j a_{j} p+j r_{j}+(p-j) a_{p-j} p+(p-j) r_{p-j}+1.$$
Because
$$j^{p}+(p-j)^{p}=\binom{p}{0} p^{p}-\binom{p}{1} p^{p-1} j+\ldots+\binom{p}{p-1} p j^{p-1}$$
we obtain that $p^{2} \mid j^{p}+(p-j)^{p}$ and we get for all $j=1,2, \ldots, p-1$,
$$j r_{j}+(p-j) r_{p-j}+1 \equiv 0 \quad(\bmod p)$$
Adding up all these relations it follows
$$2\left(r_{1}+2 r_{2}+\ldots+(p-1) r_{p-1}\right) \equiv-(p-1)(\bmod p)$$
hence
$$r_{1}+2 r_{2}+\ldots+(p-1) r_{p-1} \equiv \frac{p+1}{2}(\bmod p)$$