Problem 1313

Equivalently, $n$ is naoish iff $n = 100p - q$ where $p \ge 1$ is an integer and $1 \le q \le 10$. Decomposing each $n_j$ in this way, we have:
$$2020 = 100(p_1 + p_2 + \dots + p_k) - q_1 - q_2 - \dots - q_k.$$
In particular, $100 \mid 2020 + q_1 + q_2 + \dots + q_k$. The next multiple of $100$ above $2020$ is $2300$, so $2020 + q_1 + q_2 + \dots + q_k \ge 2300$. As each $q_j \le 10$, this implies
$$k \ge \frac{2300 - 2020}{10} = 8.$$
It is easy to find a solution that works with $k=8$. One example is $1390$ (once) and $90$ (seven times).

Let $n_i \equiv a_i \pmod{10}$ with $0 \le a_i < 10$. The second-to-last digit of $n_1 + \dots + n_k$ is the last digit of $9k + c$, where $c$ is the total carry of the addition. As $a_1 + a_2 + \dots + a_k \le 9k < 10k$ the total carry $c$ satisfies $0 \le c \le k-1$. This means that we can write
$$9k + c = 10k - (k - c) = 10k - i$$
for some $1 \le i \le k$. To get a sum of $2020$, we need to have $10k - i \equiv 2 \pmod{10}$, i.e. $i \equiv 8 \pmod{10}$ and, in particular, $i \ge 8$. From $i \le k$ we obtain $k \ge 8$.
It is easy to find a solution that works with $k = 8$. One example is $1390$ (once) and $90$ (seven times).