We claim that the answer is $\frac{(n-1)n(n+1)}{6}$. We prove this by induction. For $n=2$, the claim is obvious, since if the two numbers are $a_1 < a_2$, then we have $a_2 < a_1+a_2 < 2a_2$, which means that $a_2$ cannot divide the sum. Suppose the statement is true for some $n > 2$. Order the numbers $a_1 < a_2 < \cdots < a_{n+1}$. Using the inductive hypothesis on the first $n$ numbers, the sum of the numbers written on the board corresponding to their sums is at most $\frac{(n-1)n(n+1)}{6}$. Consider the quantities $a_{n+1}+a_i$. For each index $i$, we note that $a_i$ divides at most $k+1-i$ of these quantities, because if $a_i$ divides $a_{n+1}+a_\ell$ and $a_{n+1}+a_m$ for $\ell,m \leq i$, it also divides the difference $a_\ell-a_m$, a contradiction since $a_\ell-a_m < a_i$. For the pairs such that $a_{n+1}|a_i+a_j$, we note that $a_i+a_j < 2a_{n+1}$, and this means that $a_i+a_j=a_n$ which we ignore. Thus, for each quantity $a_{n+1}+a_i$, a number at most $n+1-i$ is written on the board. This increases the total by at most $$\sum_{i=1}^n (n+1-i)=\frac{n(n+1)}{2}$$Thus, the total is at most $$\frac{(n-1)n(n+1)}{6}+\frac{n(n+1)}{2}=\frac{n(n+1)(n+2)}{6}$$and we are done. For the construction, just take the numbers $2^1,\cdots,2^{n-1},2^n$. We note that for $i<j$, we have $2^k|2^i+2^j$ only for $k \leq i$. This makes equality hold in the above proof.