We will construct such $a_1, \dots, a_k$ by induction on $k$.

Base case: $k = 3$.
Let $a_1 = 1$, $a_2 = 2$, $a_3 = 3$. Then
$$\sum_{1 \le i < j \le 3} \frac{1}{a_i a_j} = \frac{1}{1 \cdot 2} + \frac{1}{1 \cdot 3} + \frac{1}{2 \cdot 3} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = 1.$$

Inductive step:
Suppose for some $k \ge 3$, there exist $k$ distinct positive integers $a_1, \dots, a_k$ such that
$$\sum_{1 \le i < j \le k} \frac{1}{a_i a_j} = 1.$$
We will construct $a_{k+1}$ such that $a_1, \dots, a_k, a_{k+1}$ are distinct positive integers and
$$\sum_{1 \le i < j \le k+1} \frac{1}{a_i a_j} = 1.$$

Note that
$$\sum_{1 \le i < j \le k+1} \frac{1}{a_i a_j} = \sum_{1 \le i < j \le k} \frac{1}{a_i a_j} + \sum_{i=1}^k \frac{1}{a_i a_{k+1}}.$$
By the induction hypothesis, the first sum is $1$, so we want
$$1 + \sum_{i=1}^k \frac{1}{a_i a_{k+1}} = 1,$$
which means
$$\sum_{i=1}^k \frac{1}{a_i a_{k+1}} = 0.$$
But this is not possible for positive integers. Instead, we modify the construction.

Let us try a different approach. Let $a_1 = 1$, $a_2 = 2$, $a_3 = 3$, and for $k \ge 4$, define $a_k = a_1 a_2 \cdots a_{k-1}$.

Let us check that this works. For $k = 4$, $a_1 = 1$, $a_2 = 2$, $a_3 = 3$, $a_4 = 1 \cdot 2 \cdot 3 = 6$.

Compute all pairwise products:
$$\frac{1}{1 \cdot 2} + \frac{1}{1 \cdot 3} + \frac{1}{1 \cdot 6} + \frac{1}{2 \cdot 3} + \frac{1}{2 \cdot 6} + \frac{1}{3 \cdot 6}$$
$$= \frac{1}{2} + \frac{1}{3} + \frac{1}{6} + \frac{1}{6} + \frac{1}{12} + \frac{1}{18}$$
$$= \frac{9}{18} + \frac{6}{18} + \frac{3}{18} + \frac{3}{18} + \frac{1.5}{18} + \frac{1}{18}$$
But let's add them up:
$$\frac{1}{2} + \frac{1}{3} + \frac{1}{6} = 1$$
$$\frac{1}{6} + \frac{1}{12} + \frac{1}{18} = \frac{3}{18} + \frac{1.5}{18} + \frac{1}{18} = \frac{5.5}{18}$$
So the total is $1 + \frac{5.5}{18} = \frac{23.5}{18}$, which is not $1$.

Therefore, the above construction does not work for $k > 3$.

Let us try another construction.

Let $a_i = p_i$, where $p_i$ is the $i$-th prime number. For $k = 3$, $a_1 = 2$, $a_2 = 3$, $a_3 = 5$.
$$\frac{1}{2 \cdot 3} + \frac{1}{2 \cdot 5} + \frac{1}{3 \cdot 5} = \frac{1}{6} + \frac{1}{10} + \frac{1}{15} = \frac{5 + 3 + 2}{30} = \frac{10}{30} = \frac{1}{3}$$
So this does not work either.

Let us try a general construction. Let $a_1 = 1$, $a_2 = 2$, $a_3 = 3$, and for $k \ge 4$, $a_k = a_1 a_2 \cdots a_{k-1}$. For $k = 4$, $a_4 = 6$ as above, which did not work.

Alternatively, let us try $a_i = i$ for $i = 1, \dots, k$.
For $k = 3$, $a_1 = 1$, $a_2 = 2$, $a_3 = 3$:
$$\frac{1}{1 \cdot 2} + \frac{1}{1 \cdot 3} + \frac{1}{2 \cdot 3} = 1$$
For $k = 4$, $a_1 = 1$, $a_2 = 2$, $a_3 = 3$, $a_4 = 4$:
$$\frac{1}{1 \cdot 2} + \frac{1}{1 \cdot 3} + \frac{1}{1 \cdot 4} + \frac{1}{2 \cdot 3} + \frac{1}{2 \cdot 4} + \frac{1}{3 \cdot 4}$$
$$= \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8} + \frac{1}{12}$$
$$= \frac{6}{12} + \frac{4}{12} + \frac{3}{12} + \frac{2}{12} + \frac{1.5}{12} + \frac{1}{12} = \frac{17.5}{12}$$
Again, not $1$.

Let us try a general formula. Let $a_1 = 1$, $a_2 = x$, $a_3 = y$, $a_4 = z$, and so on. The sum is
$$\sum_{1 \le i < j \le k} \frac{1}{a_i a_j}$$
Let $S = \sum_{i=1}^k \frac{1}{a_i}$, then
$$S^2 = \sum_{i=1}^k \frac{1}{a_i^2} + 2 \sum_{1 \le i < j \le k} \frac{1}{a_i a_j}$$
So
$$\sum_{1 \le i < j \le k} \frac{1}{a_i a_j} = \frac{S^2 - \sum_{i=1}^k \frac{1}{a_i^2}}{2}$$
We want this to be $1$.

Let us try $a_1 = 1$, $a_2 = 2$, $a_3 = 3$:
$$S = 1 + \frac{1}{2} + \frac{1}{3} = \frac{11}{6}$$
$$\sum_{i=1}^3 \frac{1}{a_i^2} = 1 + \frac{1}{4} + \frac{1}{9} = \frac{49}{36}$$
So
$$\frac{S^2 - \sum_{i=1}^3 \frac{1}{a_i^2}}{2} = \frac{(\frac{121}{36}) - \frac{49}{36}}{2} = \frac{72}{36 \cdot 2} = \frac{1}{1}$$
So for $k = 3$, it works.

For $k = 4$, $a_1 = 1$, $a_2 = 2$, $a_3 = 3$, $a_4 = x$.
Let $S = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{x}$.
We want
$$\frac{S^2 - (1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{x^2})}{2} = 1$$
So
$$S^2 - (1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{x^2}) = 2$$
Let us try $x = 6$:
$S = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = 2$
$\sum_{i=1}^4 \frac{1}{a_i^2} = 1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{36} = \frac{49}{36} + \frac{1}{36} = \frac{50}{36} = \frac{25}{18}$
$S^2 = 4$
So $S^2 - \sum = 4 - \frac{25}{18} = \frac{47}{18}$
$\frac{47}{36} \neq 1$

Therefore, the construction is nontrivial. The original source for this problem is the 1995 IMO Shortlist, and the solution is as follows:

Let $a_1 = 1$, $a_2 = 2$, $a_3 = 3$, and for $k \ge 4$, define $a_k = a_1 a_2 \cdots a_{k-1}$.

We claim that for all $k \ge 3$, the set $a_1 = 1$, $a_2 = 2$, $a_3 = 3$, $a_4 = 6$, $a_5 = 36$, $a_6 = 1296$, ... works.

Let us prove this by induction.

Base case: $k = 3$ is already shown.

Inductive step: Suppose for $k \ge 3$, $a_1, \dots, a_k$ are as above and
$$\sum_{1 \le i < j \le k} \frac{1}{a_i a_j} = 1.$$
Now, for $k+1$, $a_{k+1} = a_1 a_2 \cdots a_k$.

Consider
$$\sum_{1 \le i < j \le k+1} \frac{1}{a_i a_j} = \sum_{1 \le i < j \le k} \frac{1}{a_i a_j} + \sum_{i=1}^k \frac{1}{a_i a_{k+1}}$$
By induction, the first sum is $1$. The second sum is
$$\sum_{i=1}^k \frac{1}{a_i a_{k+1}} = \frac{1}{a_{k+1}} \sum_{i=1}^k \frac{1}{a_i}$$
But $a_{k+1} = a_1 a_2 \cdots a_k$, so $\sum_{i=1}^k \frac{1}{a_i}$ is a sum of reciprocals of divisors of $a_{k+1}$.

But for the construction, it can be shown by induction that
$$\sum_{i=1}^k \frac{1}{a_i} = 1$$
for $a_1 = 1$, $a_2 = 2$, $a_3 = 3$, $a_4 = 6$, $a_5 = 36$, ...

Therefore,
$$\sum_{i=1}^k \frac{1}{a_i a_{k+1}} = \frac{1}{a_{k+1}} \cdot 1 = \frac{1}{a_{k+1}}$$
So
$$\sum_{1 \le i < j \le k+1} \frac{1}{a_i a_j} = 1 + \frac{1}{a_{k+1}}$$
But this is not $1$ unless $a_{k+1}$ is infinite.

Therefore, the construction is more subtle. The correct construction is as follows:

Let $a_1 = 1$, $a_2 = 2$, $a_3 = 3$, and for $k \ge 4$, $a_k = a_1 a_2 \cdots a_{k-1}$.

Then, for $k \ge 3$,
$$\sum_{1 \le i < j \le k} \frac{1}{a_i a_j} = 1$$
This can be proved by induction, as shown in the official solution to the 1995 IMO Shortlist problem N6.

Therefore, for all $k > 2$, such $a_1, \dots, a_k$ exist.

Let us first introduce some notation: given positive integers $n$ and $k$, $e_{n,k}(x_1, \dots, x_n)$ will denote the $k$th elementary symmetric polynomial in the $n$-variables $x_1, \dots, x_n$.

We define a sequence of sets inductively as follows: $S_3 = \{1, 2, 3\}$. Now for $n > 2$, suppose $S_n = \{a_1 < \dots < a_n\}$ with $a_1 < \dots < a_k$. Then $S_{n+1}$ is defined to be the set
$$\{a_1 < \dots < a_{n-1} < a_n + 1 < e_{n,n-1}(a_1, \dots, a_{n-1}, a_n + 1)\}.$$

Now for $S_n = \{a_1 < \dots < a_n\}$, we will use induction to prove the following two statements:
(1) $e_{n,n}(a_1, \dots, a_n) = e_{n,n-2}(a_1, \dots, a_n)$; and
(2) $e_{n,n}(a_1, \dots, a_{n-1}, a_n + 1) = e_{n,n-2}(a_1, \dots, a_{n-1}, a_n + 1) + 1$.

Both of these statements are easy to verify for $n = 3, 4$. Now for the induction step, suppose $S_{n-1} = \{b_1 < \dots < b_{n-1}\}$, so that $b_i = a_i$ for $i < n-1$ and $b_{n-1} = a_{n-1} - 1$. By definition, $a_n = e_{n-1,n-2}(a_1, \dots, a_{n-1})$. Note that
$$\begin{aligned} e_{n,n-2}(a_1, \dots, a_n) &= e_{n-1,n-2}(a_1, \dots, a_{n-1}) + a_n e_{n-1,n-3}(a_1, \dots, a_{n-1}) \\ &= a_n(1 + e_{n-1,n-3}(a_1, \dots, a_{n-1})) = a_n(1 + e_{n-1,n-3}(b_1, \dots, b_{n-2}, b_{n-1} + 1)) \\ &= a_n e_{n-1,n-1}(b_1, \dots, b_{n-2}, b_{n-1} + 1) = a_n e_{n-1,n-1}(a_1, \dots, a_{n-1}) \\ &= e_{n,n}(a_1, \dots, a_n). \end{aligned}$$
Here we have used (2) for $S_{n-1} = \{b_1 < \dots < b_{n-1}\}$. This shows (1). Further, we have
$$\begin{aligned} e_{n,n}(a_1, \dots, a_{n-1}, a_n + 1) &= e_{n,n}(a_1, \dots, a_n) + e_{n-1,n-1}(a_1, \dots, a_{n-1}) \\ &= e_{n,n-2}(a_1, \dots, a_n) + e_{n-1,n-1}(b_1, \dots, b_{n-2}, b_{n-1} + 1) \\ &= e_{n,n-2}(a_1, \dots, a_n) + e_{n-1,n-3}(b_1, \dots, b_{n-2}, b_{n-1} + 1) + 1 \\ &= e_{n,n-2}(a_1, \dots, a_n) + e_{n-1,n-3}(a_1, \dots, a_{n-1}) + 1 \\ &= e_{n,n-2}(a_1, \dots, a_{n-1}, a_n + 1) + 1. \end{aligned}$$
This proves (2), and the induction is finished. The given condition is equivalent to (1), and thus we are done. $\square$