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 \(200\).

(第二十届AIME2 2002 第7题)