1. Restate the problem in a general form:
Let $a_1, a_2, \ldots, a_n$ be distinct positive integers such that $1 \leq a_1 < a_2 < \cdots < a_n < 2n$ and the least common multiple (LCM) of any two of them is $\geq 2n$. We need to show that $a_1 \geq \left\lfloor \frac{2n}{3} \right\rfloor$.

2. Initial observation:
Note that if $a_i \mid a_j$ for any $i \neq j$, then $\text{LCM}(a_i, a_j) = a_j < 2n$, which contradicts the given condition. Therefore, no $a_i$ can be a divisor of another $a_j$.

3. **Consider sequences of the form $2^k \cdot i$:**
We consider sequences $A_1, A_3, \ldots, A_{2n-1}$, where $A_i$ contains all positive integers less than $2n$ of the form $2^k \cdot i$ with $i$ being an odd number and $k$ a non-negative integer. These sequences cover all integers from $1$ to $2n-1$.

4. **Distribution of $a_i$ in sequences:**
There are exactly $n$ such sequences. If there exist distinct $a_i$ and $a_j$ in the same sequence, one must divide the other, which is a contradiction. Hence, each sequence $A_i$ contains exactly one $a_i$.

5. **Odd numbers greater than or equal to $n$:**
The sequences $A_k$ where $k$ is an odd number $\geq n$ have only one element each. Therefore, the set $\{a_i\}$ contains all odd numbers greater than or equal to $n$.

6. **Contradiction if $a_1 < \left\lfloor \frac{2n}{3} \right\rfloor$:**
Suppose $a_1 < \left\lfloor \frac{2n}{3} \right\rfloor$. We consider two cases:

**Case 1: $a_1$ is odd**
- Consider the sequence $3a_1, 5a_1, \ldots$. The difference between any two terms is $2a_1$, so any $a_1$ consecutive odd numbers contain one of $3a_1, 5a_1, \ldots$.
- If $a_1 < \frac{n}{2}$, there are at least $\frac{n}{2}$ odd numbers from $n$ to $2n$. Thus, there exists some $a_j$ such that $a_j$ is an odd multiple of $a_1$, which is a contradiction.
- If $a_1 \geq \frac{n}{2}$, then $3a_1 \geq \frac{3n}{2}$, which lies between $n$ and $2n$, leading to another contradiction.

**Case 2: $a_1$ is even**
- Let $a_1 = 2^k \cdot m$ where $m$ is an odd integer. Note that $m \leq \frac{n}{3}$.
- Consider the element $a_j$ in the sequence $A_{3m}$, and let $a_j = 2^t \cdot 3m$.
- If $k > t$, then $\text{LCM}(a_1, a_j) = 2^k \cdot 3m = 3a_j < 2n$. If $k \leq t$, then $\text{LCM}(a_1, a_j) = a_j < 2n$. Both cases yield a contradiction.

7. Conclusion:
Hence, $a_1 \geq \left\lfloor \frac{2n}{3} \right\rfloor$.

The final answer is $\boxed{ a_1 \geq 1334 }$.