We first note that there cannot exist distinct $a_i,a_j$ such that $a_i | a_j$, otherwise their LCM is less than $2n$. We consider the sequences $A_1,A_3,\cdots,A_{2n-1}$, where $A_i$ contains all positive integers $<2n$ of the form $2^k \cdot i$ where $i$ is an odd number and $k$ is a non-negative integer. Note that the elements in these sequences necessarily cover all $2n-1$ positive integers from $1$ to $2n-1$. There are exactly $n$ such sequences, and if there exist distinct $a_i, a_j$ in the same sequence, the one must divide the other, yielding a contradiction. Hence, each $A_i$ contains exactly 1 $a_i$. In particular, the sequences $A_k$ where $k$ is an odd number $\geq n$ have only one element each, hence ${a_i}$ contains all of the odd numbers greater than or equal to $n$. Now, suppose for the sake of contradiction that $a_1 < \left \lfloor \frac{2n}{3} \right \rfloor$. We consider 2 cases: Case 1: $a_1$ is odd If we look at the sequence $3a_1,5a_1,\cdots$, i.e. the odd multiplea of $a_1$, the difference between any 2 is $2a_1$ and hence any $a_1$ consecutive odd numbers contains one of $3a_1,5a_1,\cdots$. In particular, if $a_1 < \frac{n}{2}$, then since there are at least $\frac{n}{2}$ odd numbers from $n$ to $2n$, there exists some $a_j$ such that $a_j$ is an odd multiple of $a_1$, contradiction. If $a_1 \geq \frac{n}{2}$, then $3a_1$ is at least $\frac{3n}{2}$ which lies between $n$ and $2n$, which is another contradiction. Case 2: $a_1$ is even Let $a_1=2^k \cdot m$ where $m$ is an odd integer. Note that $m \leq \frac{n}{3}$. Since $m$ is odd, we consider the element $a_j$ in the sequence $A_{3m}$, and let $a_j=2^t \cdot 3m$. If $k > t$, then $lcm(a_1,a_j) = 2^k \cdot 3m = 3a_j < 2n$. If $k \leq t$, then $lcm(a_1,a_j) = a_j < 2n$. Both yield a contradiction, which finishes the case of $a_1$ even. Hence, $a_1 \ge \left \lfloor \frac{2n}{3} \right \rfloor$.