The answer is $\boxed{5}$. Claim: There exists some $k\leq 2024$ for which $a_k \leq 3$. Proof: Assume the contrary. Suppose that $a_k = p$ for some $k\leq 2023$ and some prime $p > 3$. Then, note that $a_{k+1}$ is the greatest prime factor of $a_k^2 -1 = (p-1)(p+1).$ Since $p$ is odd, we have $a_k^2-1 = 4\left(\frac{p-1}{2}\right)\left(\frac{p+1}{2}\right)$, so $a_{k+1} \leq \frac{p+1}{2} = \frac{a_k+1}{2}.$ Now, note that $a_2$ is the greatest prime factor of $a_1^2 -1 = (a_1-1)(a_1+1) = 2^{2024}(2^{2024}+2) = 2^{2025}(2^{2023}+1),$ so $a_2 \leq 2^{2023}+1$. Applying the above logic to $a_2$, we get that $a_3 \leq \frac{a_2+1}{2} \leq \frac{2^{2023}+2}{2} = 2^{2022}+1$. Similarly, $a_4 \leq 2^{2021} + 1$, $\cdots$, until we get $a_{2024} \leq 2^1+1 = 3$, a contradiction. Thus, there must exist such a $k$. $\square$ Since there exists such a $k$, note that if $a_k = 2$, then $a_{k+1} = 3$, and then $a_{k+2} = 2$, etc., and if $a_k = 3$ instead, then $a_{k+1} = 2$, $a_{k+2} = 3$, etc. So, since $k\leq 2024$, it follows that $(a_{2024}, a_{2025}) = (2, 3)$ or $(a_{2024}, a_{2025}) = (3, 2)$, so either way $a_{2024} + a_{2025} = 5$, as claimed. $\blacksquare$