We have given a positive integer $n$ with $k \geq 7$ divisors $(1) \;\; 1 = d_1 < d_2 < \cdots < d_k = n$, s.t. $(2) \;\; d_7 = 3d_4 - 1$, $(3) \;\; d_7 = 2d_5 + 1$. Claim: The smallest integer satisfying condition (1) and formulas (2)-(3) is 2024. Proof: Combining formulas (2) and (3) we obtain $3d_4 - 2d_5 = 2$, which solution is $(4) \;\; (d_4,d_5) = (2t,3t-1), t \in \mathbb{N}$. By setting $d_4=2t$ in formula (2) we find that $(5) \;\; d_7 = 6t - 1$. The fact that $d_4 = 2t \geq 4$ yields $t \geq 2$. Assume $t=2$. Then $(d_4,d_5,d_7) = (4,5,11)$ by formulas (4)-(5), yielding $d_6 = 2 \cdot 3 = 6$ and $d_7 \leq 2 \cdot 5 = 10$, contradicting $d_7=11$. Consequently $t \geq 3$, which implies (by condition (1) and the fact that $d_4=2t$) $(6) \;\; (d_2,d_3) = (2,t)$. Next assume $t=3$. Then $(d_4,d_5)=(6,8)$ by formulas (4), yielding $(d_3,d_4)=(3,4)$, which is impossible since $d_4=6$. Therefore $t \geq 4$. Next assume $t=4$. Then by formulas (4)-(6) we obtain $(d_2,d_3,d_4,d_5,d_7) = (2,4,8,11,23)$, which means $d_2 \cdot d_5 = 2 \cdot 11 = 22$ is a divisor of $n$, yielding $d_6=22$ (since $d_5=11 < d_6 < 23=d_6$ by condition (1)). Checking we find that $n = d_3 \cdot d_5 \cdot d_7 = 8 \cdot 11 \cdot 23 =2024$ satisfies condition (1) and formulas (2)-(3). The fact that $(d_2,d_5,d_7) = (t,3t-1,6t-1)$ is a triple of three mutally coprime distinct divisors of $n$ implies $d_2d_5d_7$ divides $n$, i.e. $(7) \;\; P(t) = t(3t - 1)(6t - 1) \mid n$. The fact that $2024 = 2P(4)=2024 < 2030 = P(5) \leq P(t)$ for all $t>4$ completes the proof.