http://www.kalva.demon.co.uk/short/soln/sh901.html

Define $n$ such a integer, if it exists; then by assumption there exists a integer $m$ such that $n=\sum_{i=0}^{1989}{m+i}=5\cdot 199 (2m+1989)$, and in particular $n$ is odd. The other assumption says that $2n$ can be expressed as $(k+1)(2n+k)$ in exactly $1990$ ways: but if $2n=ab$ for some integers $1<a<b$ then the solution is unique $(n,k)=(\frac{1}{2}(b+1-a),a-1)$, i.e. there are exactly $\frac{1}{2}\sigma_0(2n)-1$ such factorization. It implied that $\sigma_0(2n)=3982$. But $2n=2\cdot 5^{e_2}\cdot 199^{e_3} \cdot \prod_{i=4}^{\omega}{p_i^{e_i}}$ with the usual notation, and $\sigma_0(2n)=2\prod_{i=1}^\omega{(e_i+1)}=11\cdot 181$. By force $\{e_1,e_2\}=\{10,180\}$, so that we have exactly two solutions: $n_1=5^{10}\cdot 199^{180}$ and $n_2=5^{180}\cdot 199^{10}$. []