Answer: $\frac{d}{2}$

As $n > 1$ is squarefree, it can't be a square of an integer. Then all divisors of $n$ can be split into pairs $(t_1, t_2)$, $(t_3, t_4)$, ..., $(t_{d-1}, t_d)$ in such a way, that the product of numbers in each pair is $n$. If we choose $(a, b)$ from the same pair, then $a^2 + ab - n = a^2$, contradiction. So, we chose at most one number from each pair, so we selected at most $\frac{d}{2}$ in total.

Let's show that we can always choose $\frac{d}{2}$ integers. Consider any prime divisor $p$ of $n$ and choose all divisors of $n$, which are divisible by $p$. It's easy to see that there are $d/2$ of them. Let's show that they satisfy the condition from the statement. Let $a = kp, b = lp$ be any divisors from the chosen group, and $n = pt$. Note that $l, k, t$ aren't divisible by $p$. Then $a^2 + ab - n = p(pk^2 + pkl - t)$ and the number in brackets isn't divisible by $p$, as $t$ isn't divisible by $p$. So, $a^2 + ab - n$ is divisible by $p$, but not by $p^2$, and, therefore, can't be a square of an integer.