Suppose $a_1 = 2^0, a_2 = 2^1, \dots, a_{n-2} = 2^{n-3}, a_{n-1} = 3, a_n = q$. We shall prove there exists infinitely many positive integers $n$ for which there exists a positive integer $q$ such that $(q, 6) = 1$ and $a_1, \dots, a_n$ are harmonic.

$$\begin{align*} \sum_{1 \le i < j \le n} (a_i, a_j) &= \sum_{1 \le i < j \le n-2} (2^{i-1}, 2^{j-1}) + \sum_{1 \le i \le n-2} (2^{i-1}, 3) + \sum_{1 \le i \le n-2} (2^{i-1}, q) + (3, q) \\ &= \sum_{1 \le i < j \le n-2} 2^{i-1} + n - 2 + n - 2 + (3, q) \end{align*}$$

$a_1, a_2, \dots, a_n$ is harmonic if and only if

$$\begin{align*} \sum_{i=1}^{n} a_i &= \sum_{1 \le i < j \le n} (a_i, a_j) \\ \iff 2^{n-2} - 1 + 3 + q &= \sum_{1 \le i \le n-2} (n-2-i)2^{i-1} + 2n-3 \\ \iff q &= \sum_{2 \le i \le n-2} (n-2-i)2^{i-1} + (n-3) + 2n-3 - 2^{n-2} + 2 \\ &= \sum_{2 \le i \le n-2} (n-2-i)2^{i-1} - 2^{n-2} + 3n-4 \\ &= \sum_{0 \le i \le n-3} (n-3-i)2^i - 2^{n-2} + 2n-1 = A(n) \end{align*}$$

$$\begin{align*} \sum_{0 \le i \le n-3} (n-3-i)2^i &= \sum_{0 \le i \le n-3} \sum_{0 \le j < i} 2^j \\ &= \sum_{0 \le i \le n-3} (2^i - 1) = 2^{n-2} - 1 - (n-2) \\ \implies A(n) &= -1 - (n-2) + 2n-1 = n \implies q = n \end{align*}$$

Therefore, for every positive integer $n$ that $(n, 6) = 1$, the sequence $a_1, \dots, a_n$ will be harmonic. ■