Peter wrote: Let $a\ge0$ be the smallest integer such that $a^{2}+a+n$ is composite, having smallest prime divisor $b$. Assume for the sake of contradiction that $a\le n-2$. If $b\ge 2a+1$, we have $a^{2}+a+n>a^{2}>b^{2}\be(2a+1)^{2}$, so $n>3a^{2}$ so $a<\sqrt{n/3}$, contradiction. So $b<2a+1$, then since $(a^{2}+a+n)-(c^{2}+c+n)=(a-c)(a+c+1)$, there is a positive integer $c<a$ for which one factor equals $b$. Now $b|(a^{2}+a+n)-(c^{2}+c+n)$ and $b|a^{2}+a+n$ so $b|c^{2}+c+n$, since $c<a$ this means $c^{2}+c+n=b$. Now an easy estimation gives $a-c<a+c+1<(n-2)+c+1<c^{2}+c+n=b$, thus $b$ is prime but $b$ can equal none of these factors, contradiction.

Just a side note: Using heavier artillery, we can show that the requirements are fulfilled for $ n=2, 3, 5, 11, 17, 41$ only.