USA IMO TST

The answer is $0 \le r \le 1/2$. Obviously $r \ge 0$. In one direction, we show that
Claim (Greedy bound) — For all integers $n$, we have
$$\frac{b_n}{n^2} \le \frac{1}{2} + \frac{1}{2n}.$$
Proof. This is by induction on $n$. For $n=1$ it is given. For the inductive step we have
$$\begin{align*} b_n &< n^2 \frac{b_{n-1}}{(n-1)^2} \\ &\le n^2 \left( \frac{1}{2} + \frac{1}{2(n-1)} \right) = \frac{n^3}{2(n-1)} \\ &= \frac{1}{2} \left[ n^2 + n + 1 + \frac{1}{n-1} \right] \\ &= \frac{n(n+1)}{2} + \frac{1}{2} \left[ 1 + \frac{1}{n-1} \right] \\ &\le \frac{n(n+1)}{2} + 1 \end{align*}$$
So $b_n < \frac{n(n+1)}{2} + 1$ and since $b_n$ is an integer, $b_n \le \frac{n(n+1)}{2}$. This implies the result. $\square$

We now give a construction. For $r = 1/2$ we take $b_n = \frac{1}{2}n(n+1)$; for $r = 0$ we take $b_n = 1$.

Claim (Explicit construction, given by Nikolai Beluhov) — Fix $0 < r < 1/2$. Let $N$ be large enough that $\lceil rn^2 + n \rceil < \frac{1}{2}n(n+1)$ for all $n \ge N$. Then the following sequence works:
$$b_n = \begin{cases} \lceil rn^2 + n \rceil & n \ge N \\ \frac{n^2+n}{2} & n < N. \end{cases}$$
Proof. We certainly have
$$\frac{b_n}{n^2} = \frac{rn^2 + n + O(1)}{n^2} \xrightarrow{n \to \infty} r.$$
Mainly, we contend $b_n n^{-2}$ is strictly decreasing. We need only check this for $n \ge N$; in fact
$$\frac{b_n}{n^2} \ge \frac{rn^2 + n}{n^2} > \frac{[r(n+1)^2 + (n+1)] + 1}{(n+1)^2} > \frac{b_{n+1}}{(n+1)^2}$$
where the middle inequality is true since it rearranges to $\frac{1}{n} > \frac{n+2}{(n+1)^2}$. $\square$