Denote by $R(n)$ the number of red numbers in $\{1,2, \ldots, n\}$, i.e., $R(n)=n r(n)$. Similarly, denote by $B(n)$ and $b(n)=B(n) / n$ the number and proportion of blue numbers in $\{1,2, \ldots, n\}$, respectively. Notice that $B(n)+R(n)=n$ and $b(n)+r(n)=1$. Therefore, the statement of the problem does not change after swapping the colors. Arguing indirectly, for every $a \in\{1,2, \ldots, p-1\}$ choose some positive integer $n_{a}$ such that $r\left(n_{a}\right)=a / p$ and, hence, $R\left(n_{a}\right)=a n_{a} / p$. Clearly, $p \mid n_{a}$, so that $n_{a}=p m_{a}$ for some positive integer $m_{a}$, and $R\left(n_{a}\right)=a m_{a}$. Without loss of generality, we assume that $m_{1}m_{p-1}$. Choose $k$ to be the smallest index satisfying $m_{k}>m_{p-1}$; by our assumptions, we have $1<k \leqslant q<p-1$. Let $m_{a}$ be the maximal number among $m_{1}, m_{2}, \ldots, m_{k-1}$; then $a \leqslant k-1 \leqslant m_{a}<m_{p-1}$. Applying the Claim, we get $$\begin{aligned} m_{k} \geqslant \frac{p-1}{k} m_{p-1} \geqslant \frac{p-1}{k} & \cdot \frac{p-a}{p-(p-1)} m_{a} \\ & \geqslant \frac{p-1}{k} \cdot(p-k+1)(k-1) \geqslant \frac{k-1}{k} \cdot(p-1)(p-q) \geqslant \frac{1}{2} \cdot \frac{p^{2}-1}{2} \end{aligned}$$ which contradicts (1) again. Comment 1. The argument in Case 2, after a slight modification of estimates at the end, applies as soon as there exists $k<\frac{3(p+1)}{4}$ with $a_{k}<a_{p-1}$. However, this argument does not seem to work if there is no such $k$. Comment 2. If $p$ is small enough, then one can color $\{1,2, \ldots, N+1\}$ so that there exist numbers $m_{1}$, $m_{2}, \ldots, m_{p-1}$ satisfying $r\left(p m_{a}\right)=a / p$. For $p=3,5,7$, one can find colorings providing the following sequences: $$\left(m_{1}, m_{2}\right)=(1,2), \quad\left(m_{1}, m_{2}, m_{3}, m_{4}\right)=(1,2,3,6), \quad \text { and } \quad\left(m_{1}, \ldots, m_{6}\right)=(1,2,3,4,6,12)$$ respectively. Thus, for small values of $p$, the number $N$ in the problem statement cannot be increased. However, a careful analysis of the estimates shows that this number can be slightly increased for $p \geqslant 11$.