Balkan Mathematical Olympiad competition problems

The answer is $n = \lceil \log_{K+1} N \rceil$.

Note first that for $n \le \lceil \log_{K+1} N \rceil$, taking $a_i = (K+1)^{i-1}$ works. Indeed let $r$ be maximal such that $m_r \ne 0$. Then on the one hand we have
$$\left| \sum_{i=1}^{n} m_i a_i \right| \le \sum_{i=1}^{n} K (K+1)^{i-1} = (K+1)^n - 1 < N.$$
On the other hand we have
$$\left| \sum_{i=1}^{n} m_i a_i \right| \ge |m_r a_r| - \left| \sum_{i=1}^{r-1} m_i a_i \right| \ge (K+1)^{r-1} - \sum_{i=1}^{r-1} K (K+1)^{i-1} = 1 > 0.$$
So the sum is indeed not divisible by $N$.

Assume now that $n \ge \lceil \log_{K+1} N \rceil$ and look at all $n$-tuples of the form $(t_1, \dots, t_n)$ where each $t_i$ is a non-negative integer with $t_i \le K$. There are $(K+1)^n > N$ such tuples so there are two of them, say $(t_1, \dots, t_n)$ and $(t'_1, \dots, t'_n)$ such that
$$\sum_{i=1}^{n} t_i a_i \equiv \sum_{i=1}^{n} t'_i a_i \pmod N.$$
Now taking $m_i = t_i - t'_i$ for each $i$ satisfies the requirements on the $m_i$'s but $N$ divides the sum
$$\sum_{i=1}^{n} m_i a_i,$$
a contradiction.