By Fermat's Little Theorem, we have $j^{p-1} \equiv 1 \mod p \implies j^{p-1} - 1 \equiv 0 \mod p$. In addition, we also have $$\frac{j^{p-1}-1}{p} \equiv r_j \mod p$$ $$j^{p-1} - 1 \equiv pr_j \mod p$$ $$j^{p-1} - 1 \equiv r_j \mod p$$ However, from our first conclusion we can see that $r_j \equiv 0 \mod p$, which collapses the sum, makes $jr_j \equiv 0 \mod p$ and makes the sum $$S \equiv 0 \equiv \frac{p+1}{2} \mod p$$or $$p \equiv -1 \mod p$$ which is a contradiction to the problem. Hence our original statement is not true?

$jr_j+ (p - j)r_{p-j}+1 \equiv 0 \, (\mod p)$ for $j=1,2,..., p-1$

Crucial Claim. We have $jr_j + (p-j)r_{p-j} \equiv -1 \mod p$ for all $1 \le j \le p-1$. Proof: We may write $r_j \equiv \frac{j^{p-1} - 1}{p} \mod p$ and $r_{p-j} \equiv \frac{(p-j)^{p-1} - 1}{p} \mod p$. Multiplying by $j$ and $p-j$ respectively gives $$jr_j \equiv \frac{j^{p}-j}{p} \mod p$$and $$(p-j)r_{p-j} \equiv \frac{(p-j)(p-j)^{p-1} - (p-j)}{p} = \frac{(p-j)^{p} + j - p}{p} \mod p$$. Adding these two congruences together will yield $$jr_j + (p-j)r_{p-j} \equiv \frac{j^{p}-j + (p-j)^p + j - p}{p} \mod p$$ We must further note that by the Binomial Theorem, $$(p-j)^p = p^p + \binom{p}{1}p^{p-1}(-j)^1 \binom{p}{2}p^{p-2}(-j)^2 + \cdots + \binom{p}{p-1}p^{1}(-j)^{p-1} + (-j)^{p}$$ This is important because we see that the first $p-1$ terms are divisible by $p$. Also, $(-j)^{p} = j^{p}$ since $p$ is a prime and odd. Hence because $j<p$, we can say that $(p-j)^p \equiv -j^{p} \mod p$. Our modulus will cancel out: $$jr_j + (p-j)r_{p-j} \equiv \frac{j^{p}-j + (p-j)^p + j - p}{p} \mod p$$$$\equiv \frac{j^{p}-j-j^{p}+j-p}{p} \mod p$$$$\equiv \frac{-p}{p} = \boxed{-1 \mod p}$$Our crucial claim is therefore true. Continuing, we must see that if we sum over $1 \le j \le p$, we will obtain: $$\sum_{j=1}^{p-1} jp_j + (p-j)r_{p-j} = 2(r_1+2r_2+\cdots + (p-1)r_{p-j}).$$Since we are iterating over this range, there are $p-1$ sums that will contribute a $-1$ when we take this sum modulus $p$. Therefore, we have our desired expression, $$2(r_1+2r_2+\cdots + (p-1)r_{p-j}) \equiv p-1 \mod p$$ $$\implies r_1+2r_2+\cdots + (p-1)r_{p-j} \equiv \frac{p-1}{2} \equiv \frac{p+1}{2} \mod p$$ Note that division by $2$ above is legal because of the fact that $p$ is a prime greater than or equal to $3$, which implies that $p$ is odd and relatively prime to $2$. Remarks. I really liked solving this problem! Although this is a number theory olympiad problem and that my number theory is not very strong, it still taught me a lot about how collapsing a sum using symmetry can help a lot. Also, the binomial theorem can help a lot with number theory problems