jmc

$\frac{k(k+1)(2k+1)}{6}$ is a multiple of $200$ if $k(k+1)(2k+1)$ is a multiple of $1200=2^4\cdot 3\cdot 5^2$. So $16,3,25|k(k+1)(2k+1)$. Since $2k+1$ is always odd, and only one of $k$ and $k+1$ is even, either $k,k+1\equiv 0(\mathrm{mod}16)$. Thus, $k\equiv 0,15(\mathrm{mod}16)$. If $k\equiv 0(\mathrm{mod}3)$, then $3|k$. If $k\equiv 1(\mathrm{mod}3)$, then $3|2k+1$. If $k\equiv 2(\mathrm{mod}3)$, then $3|k+1$. Thus, there are no restrictions on $k$ in $(\mathrm{mod}3)$. It is easy to see that only one of $k$, $k+1$, and $2k+1$ is divisible by $5$. So either $k,k+1,2k+1\equiv 0(\mathrm{mod}25)$. Thus, $k\equiv 0,24,12(\mathrm{mod}25)$. From the Chinese Remainder Theorem, $k\equiv 0,112,224,175,287,399(\mathrm{mod}400)$. Thus, the smallest positive integer $k$ is $\boxed{112}$.