We argue a contradiction. Assume such a partition exists. Then, it holds that for some $k,\ell,m\in\mathbb{N}$ \[ 1+2+\cdots+1998 = 999\cdot 1999 = 2000k+3999\ell+5998 m. \]Inspecting both sides modulo $1999$, we find that $1999\mid k+\ell+m$. Now, if $k\ge 998$, then since \[ \frac{999}{998}>\frac{2000}{1999} \]we have a contradiction. Assume therefore that $k\le 997$, thus $\ell+m\ge 1999-997 = 1002$. Hence, $\max\{\ell,m\}\ge 501$. With this, we obtain \[ 2000k+3999\ell+5998m >501\cdot 3999 >1002\cdot 1999>999\cdot 1999. \]These chain of inequalities yields a contradiction.