A beautiful problem , even though i found it hard . First, note that $n,n+1, \dots \ 2n-1$ give different residue $\pmod {n+i}$ for $i \in \{0,1,2,\dots,n-1 \}$ Let $S$ be the sum of all numbers on the blackboard , $S=n^2+\frac{n(n-1)}{2}=\frac{n(3n-1)}{2}$ If we erase any number $x$ the sum will be $S-x$ . If two of $S-x$'s are divided by the same number then $${S-x} \equiv {S-y} \pmod{n+i} \Longleftrightarrow x \equiv y \pmod{n+i}$$Wich is absurd , so all of $n-1$ , $S-x$'s have different divisors from the set of written numbers(on blackboard). Assume that all $S-x$'s are divided by a written number , then all of them must be divided by exactly one of the written numbers (because there are $n-1$ numbers and $n-1$ sums) But taking $x=n$ the sum is $\frac{n(3n-3)}{2}$ wich is divided by both $n$ and $\frac{3n-3}{2}$ (because $n$ is odd and $n <\frac{3n-3}{2}<2n-1$ is a written integer .Contradiction .$\blacksquare$