6. 16 .

From the given conditions, we have
$$\begin{array}{l} f(x)=\left(\sin ^{2} \frac{k x}{10}+\cos ^{2} \frac{k x}{10}\right)^{2}-2 \sin ^{2} \frac{k x}{10} \cdot \cos ^{2} \frac{k x}{10} \\ =1-\frac{1}{2} \sin ^{2} \frac{k x}{5}=\frac{1}{4} \cos \frac{2 k x}{5}+\frac{3}{4}, \end{array}$$

The function $f(x)$ reaches its maximum value if and only if $x=\frac{5 m \pi}{k}(m \in \mathbf{Z})$.

According to the conditions, we know that any open interval of length 1, $(a, a+1)$, contains at least one maximum point. Therefore,
$$\frac{5 \pi}{k}5 \pi \text {. }$$

Conversely, when $k>5 \pi$, any open interval $(a, a+1)$ contains a complete period of $f(x)$, at which point, $\{f(x) \mid a<x<a+1\}=\{f(x) \mid x \in \mathbf{R}\}$.
In summary, the minimum value of $k$ is $[5 \pi]+1=16$, where $[x]$ denotes the greatest integer not exceeding the real number $x$.