Let $M_k$ denote the move that increases the blue number by $k+1$ and the red number by $k^2+2$. We show that for any $k \ge 2$, the move $M_k$ can be replaced by moves involving only $M_0$ and $M_1$ in such a way that the total increase in the blue number is the same, but the red number increases by less than $k^2+2$.

Case 1: $k = 2p + 1$ is odd. Then $M_k$ increases the blue number by $2p + 2$ and the red number by $(2p+1)^2 + 2$. Instead, $p+1$ moves of type $M_1$ increase the blue number also by $2p+2$, but the red number increases by $3(p+1)$. Since $(2p+1)^2 + 2 > 3(p+1)$, for all $p \ge 1$, we can replace $M_k$ with $p+1$ moves of $M_1$.

Case 2: $k = 2p$ is even. Then $M_k$ increases the blue number by $2p+1$, and the red number by $(2p)^2 + 2$. Alternatively, $p$ moves of $M_1$ and one move of $M_0$ increase the blue number by $2p+1$ and the red number by $3p+2$. Since $(2p)^2 + 2 > 3p+2$, for all $p \ge 1$, we conclude that $M_k$ can be replaced by $p$ moves of $M_1$ and one $M_0$.
Moreover, observe that two moves of $M_0$ can be replaced by one move of $M_1$, with a smaller increase in the red number: $3$ instead of $4$.

Thus, using only $M_0$ and $M_1$, and minimizing the use of $M_0$, yields the minimal red number. To make the blue number at least $N$, the best strategy is to use only moves of type $M_1$ (which increase blue by $2$ and red by $3$), plus at most one $M_0$ (if needed).

If $N$ is even: we can use $\left\lfloor \frac{N}{2} \right\rfloor$ moves of $M_1$, increasing red by $3 \cdot \frac{N}{2} = \frac{3N}{2}$.

If $N$ is odd: we can use $\left\lceil \frac{N}{2} \right\rceil$ moves of $M_1$ and one move of $M_0$, leading to red increasing by $3\left\lfloor \frac{N}{2} \right\rfloor + 2$.

Hence, the minimal possible value of the red number is $N + \left\lfloor \frac{N+1}{2} \right\rfloor$.