10.3. The function $f(x)=\prod_{i=1}^{2000} \cos \frac{x}{i}$ changes sign how many times in the interval $\left[0, \frac{2009 \pi}{2}\right]$?
10.3. 75 times.

Let $n=2009$. Consider the function $\cos \frac{x}{k}$. It changes sign at $x=\frac{k(2 m+1) \pi}{2}$. This indicates that the zeros of $f(x)$ are $x_{i}=\frac{i \pi}{2}(1 \leqslant i \leqslant n)$.

We only need to consider the sign change of $f(x)$ at $x_{i}(i=1,2, \cdots, n-1)$.
$\cos \frac{x}{k}$ changes sign at $x_{i}$ if and only if $i=k(2 m+1)$.
The number of functions $\cos x, \cos \frac{x}{2}, \cdots, \cos \frac{x}{n}$ that change sign at $x_{i}$ is equal to the number of odd divisors of $i$. Therefore, $f(x)$ changes sign at $x_{i}$ if and only if $i$ has an odd number of odd divisors.

Let $i=2^{l} j$ (where $j$ is an odd number). Then $i$ and $j$ have the same number of odd divisors. An odd number has an odd number of odd divisors if and only if

it is a perfect square, $j$ is a perfect square if and only if $2^{l} j$ is a perfect square or twice a perfect square (depending on the parity of $l$), there are $[\sqrt{n-1}]$ perfect squares and $\left[\sqrt{\frac{n-1}{2}}\right]$ double perfect squares in the range $1 \sim n-1$.
Thus, the number of sign changes is
$$
[\sqrt{n-1}]+\left[\sqrt{\frac{n-1}{2}}\right]=44+31=75 .
$$