Such a 'large' set must have infinitely many distinct pairs that differ by perfect squares.
This follows from a quantitative strengthening of Sárközy's theorem proved by Pintz-Steiger-Szemerédi. The same is true if we replace squares by any other power (so perfect cubes, fourth powers, and so on).
In general, the following is true. We say that $h\in\mathbb{Z}[x]$ is an intersective polynomial if it is non-zero and for every $q$ there is some $n$ such that $q\mid h(n)$ (for example, $h(x)=x^k$ is intersective). If $A\subset \mathbb{N}$ is such that $\sum_{a\in A}\frac{1}{a}=\infty$ then for any fixed intersective polynomial $h$ there are infinitely many pairs $a\neq b\in A$ such that $a-b=h(n)$ for some $n$. This generalisation of the Pintz-Steiger-Szemerédi result was proved by Alex Rice (https://arxiv.org/pdf/1612.01760.pdf) (the case when $h$ is a perfect power was already proved by Balog-Pelikan-Pintz-Szemerédi).
Note being intersective is also necessary for such a property to hold, otherwise if $q$ is such that $h(n)\equiv 0\pmod{q}$ has no solutions then $A$ being the set of all multiples of $q$ is a 'large' set that has no solutions to $a-b=h(n)$.
If $y_n$ is increasing, $y_n\geq n+y_0$ for all $n$, so $x_n\geq \frac{n(n-1)}{2}+ny_0 + x_0$.
$x_n$ is of quadratic growth, so the sum of reciprocals converges, thus it can't be a counter-example.
If $y_n$ is only non-decreasing, we need to have a constant $c$ that bounds the number of consecutive identical values, otherwise, we would get arbitrarily long constant sequences of $y$ values, so arbitrarily long arithmetic progressions of $x$ values.
We have:
$$x_{n+c} = \sum_{k=0}^{c-1} y_{n+k} + x_n \geq cy_{n} + x_n$$
Let $w_n = x_{cn}$.
$$w_{n+1} = x_{cn+c} \geq cy_{cn} + x_{cn} = cy_{cn} + w_n$$
But $y_{cn}$ must be strictly increasing, so we get that the sum of the reciprocals of $(x_{cn})_n$ is converging.
We have:
$$\sum_{n=0}^{+\infty} \frac{1}{x_n} \leq \sum_{n=0}^{+\infty} \frac{1}{x_{c\left\lceil \frac{n}{c} \right\rceil}} = c\sum_{k=1}^{+\infty}\frac{1}{x_{ck}}$$
Hence the convergence.
Best Answer
This is a remarkable conjecture and an example of exceptional mathematical intuition. So it is hard to say exactly how it came about.
The sum of reciprocals of an arithmetic progression of real numbers (containing no zero elements) will diverge by comparison with the harmonic series (or more easily if the common difference is zero). So one might think that there would be a connection between arithmetic progressions and divergence.
But consider, for example, $A$ as the union of arithmetic progressions of length $n$ starting at $n^3$ for which the sum of reciprocals would converge by comparison with the sum of $\frac 1{n^2}$ - arithmetic progressions of arbitrary length do not guarantee divergence - they can be very sparse.
At the time the conjecture was made it was known that the sum of the reciprocals of the primes diverged, but not that the primes contained arbitrarily long arithmetic progressions. And it is not difficult to create sums of reciprocals which diverge more slowly than the primes.
Notice how the conjecture relates to the density of $A$ within the natural numbers, and would encompass sparser sets than the primes. It represents a serious mathematical challenge.
I think the value of the conjecture is that it is both mathematically coherent - it relates things - divergence and arithmetic progressions - which have some relationship already; and it is mathematically challenging.
If you follow the information about Szemeredi's Theorem you will see how many first-rate mathematicians have contributed in this area of how sparse a set can be and yet still be guaranteed to contain arbitrarily long arithmetic progressions.
The conjectures we hear about are the ones which survive the test of time - the trivial, and the trivially wrong are soon lost to memory, and the merely difficult generate papers and solutions. Others, like this one, sit at the boundary of what is known to be accessible, and what looks to be just beyond reach. That is what makes them intriguing.