It is known that, for all positive integers $k$,
$1^2+2^2+3^2+\ldots+k^{2}=\frac{k(k+1)(2k+1)}6$.
Find the smallest positive integer $k$ such that $1^2+2^2+3^2+\ldots+k^2$ is a multiple of $200$.

Let's think step by step and output the final answer within \boxed{}.
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