Denote the numbers by $a_1, a_2, \dots, a_n$. Without loss of generality we may assume that $a_1 > a_2 > \dots > a_n$. Let us show we may also assume that not all of these numbers are divisible by $3$. If $b_1, \dots, b_n$ are all divisible by $3$, then there exists a positive integer $k$ such that $3^k$ divides $b_j$ for all $j = 1, \dots, n$, as well as a positive integer $l$ such that $3^{k+1}$ does not divide $b_l$. If this is the case, then $a_1 = \frac{b_1}{3^k}, \dots, a_n = \frac{b_n}{3^k}$ are pairwise different positive integers, which are not all divisible by $3$. Assume that there exist two among them, such that $a_i + a_j$ does not divide any of the numbers $3a_1, \dots, 3a_n$. Then $3^k(a_i + a_j) = b_i + b_j$ does not divide $3^k(3a_1) = 3b_1, \dots, 3^k(3a_n) = 3b_n$.

Let $a_1 > a_2 > \dots > a_n$ and assume $a_1, \dots, a_n$ are not all divisible by $3$. We will prove the claim by contradiction. If the claim is not true, then for every sum $a_i + a_j$ there exists an index $k_{ij}$ such that $3a_{k_{ij}}$ is a multiple of this sum. In particular, this holds for $i = 1$, so $a_i + a_1$ divides $3a_{k_{i1}}$ for some $k$. If $a_i + a_1$ is not divisible by $3$, then $a_i + a_1$ divides $a_{k_{i1}}$. This is impossible since $a_i + a_1 > a_{k_{i1}}$. So, $3$ divides $a_i + a_1$ for all $2 \le i \le n$. At least one of the numbers is not divisible by $3$ which implies none of them are and $a_2, \dots, a_n$ all give the same remainder when divided by $3$.

This remainder is non-zero, so $a_i + a_2$ is not divisible by $3$ for $i = 3, \dots, n$. Now, $a_i + a_2$ divides $3a_{k_{i2}}$, so $a_i + a_2$ divides $a_{k_{i2}}$. This is only possible if $k_{i2} = 1$. Hence, $a_3 + a_2, \dots, a_n + a_2$ all divide $a_1$.

We know that $(a_1 + a_2)l = 3a_m$ for some $m$ and some positive integer $l$. Obviously, $l \ge 3$ implies $(a_1 + a_2)l > 3a_1 > 3a_m$, so $l = 1$ or $l = 2$. If $l = 2$, then $3a_m = 2(a_1 + a_2) > 4a_2$, which would imply $m = 1$ and $a_1 = 2a_2$. We have shown that $a_2 + a_3$ divides $a_1$ and when $a_1 = 2a_2$ we have $2(a_2 + a_3) = a_1 + 2a_3 > a_1$, so $a_2 + a_3 = a_1 = 2a_2$ or $a_2 = a_3$. This is not possible since all the numbers on the blackboard are different. We conclude that $l = 1$.

We have shown that $a_1 + a_2 = 3a_m$ for some $m$. Since
$$3a_m = a_1 + a_2 \ge a_2 + a_3 + a_3 > 3a_3$$
we have $m < 3$. If $m = 1$ then $a_1 = 2a_2$, but this is not possible since $a_1 > a_2$. Hence $m = 2$ and $a_1 = 2a_2$. As above, since $2(a_2 + a_3) = a_1 + 2a_3 > a_1$ and $a_2 + a_3$ divides $a_1$, we have $a_2 + a_3 = a_1 = 2a_2$ or $a_2 = a_3$. This contradicts the assumption that all numbers are different.

We have arrived at a contradiction and we can conclude that it is always possible to choose two of the numbers so that any other number from the board multiplied by $3$ is different from their sum.