Suppose, on the contrary, that the function $f$ is periodic with the period $T$ for some integer $n \ge 2$. Hence, $f(T) = f(0) = n$.
Now we have
$$f(T) = \cos(T\sqrt{1}) + \cos(T\sqrt{2}) + \dots + \cos(T\sqrt{n}) = n,$$
from which we conclude that $\cos(T\sqrt{1}) = \cos(T\sqrt{2}) = \dots = \cos(T\sqrt{n}) = 1$. So, $T = 2k\pi$ and $T\sqrt{2} = 2l\pi$, where $k, l \in \mathbb{N}$. Hence, $\sqrt{2} = l/k \in \mathbb{Q}$, which is contradiction. In conclusion, there is no integer $n \ge 2$ such that $f$ is periodic.