We will prove that $n$ is a "good" number if and only if it is prime.

First we will prove that if $n$ is not prime, then it is not "good". Let $n = rs$, $1 < r, s < n$. For $b_1 = b_2 = \dots = b_{n-1} = r$ then
$$\{\sum b_i \pmod n \mid I \subseteq \{1, 2, \dots, n-1\}\} = \{0, r, 2r, \dots, n - r\}$$
and for $i = 1$ there does not exist a subset $I$.

Now we prove that every prime number is "good".
Let $p$ be a prime number. There exist at least $r + 1$ different numbers (mod $p$) as sums of the elements $b_1, b_2, \dots, b_r$, where $1 \le r \le n-1$, $1 \le b_1, b_2, \dots, b_r \le p-1$. The proof is by induction.

For $r = 1$, empty sum is equal to zero (mod $p$) and the sum of $b_1$ is not.

Let $r < n-1$ and let the proposition be true for $r$.
Let the proposition be not true for $p + 1$. Let $1 \le b_1, b_2, \dots, b_r, b \le p-1$ be the numbers such that their sums do not give at least $r + 2$ different numbers (mod $p$). Since the proposition holds for $r$, there exist sums $0 = \sigma_0, \sigma_1, \dots, \sigma_r$ distinct from each other (mod $p$), for $b_1, b_2, \dots, b_r$. Then $\sigma_0 + b, \sigma_1 + b, \dots, \sigma_r + b$ do not give a new number (mod $p$) (different from $0 = \sigma_0, \sigma_1, \dots, \sigma_r$). Because $\sigma_i + b \ne \sigma_i \pmod p$, it holds $0, b, 2b, \dots, (r+1)b \in \{\sigma_0, \sigma_1, \dots, \sigma_r\}$. But there exist $i, j$ such that $0 \le i < j \le r+1$ and $ib \equiv jb \pmod p$, from where $(j-i)b \equiv 0 \pmod p$. This is a contradiction with $p$ being a prime number.