4. Let the function $y=f(x)$ have a domain of $\mathbf{R}$, and for $x>1$, and for any real numbers $x, y \in \mathbf{R}$, $f(x+y)=f(x) f(y)$ holds. The sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=f(0)$, and $f\left(a_{n+1}\right)=\frac{1}{f\left(-2-a_{n}\right)}(n \in \mathbf{N})$. If the inequality $\left(1+\frac{1}{a_{1}}\right)\left(1+\frac{1}{a_{2}}\right) \cdots\left(1+\frac{1}{a_{n}}\right) \geqslant k \cdot$ $\sqrt{2 n+1}$ holds for all $n \in \mathbf{N}$, find the minimum value of $k$. (2003 Hunan Province Mathematics Competition Problem)
4. Let $x=-1, y=0$, we get
$$f(-1)=f(-1) f(0)=f(0) \equiv 1$$

Thus $a_{1}=f(0)=1$.
When $x>0$, $-x<0$
\end{aligned}$$

Therefore, $f\left(x_{1}\right)>f\left(x_{2}\right)$, the function $y=f(x)$ is monotonically decreasing on $\mathbf{R}$.
From $f\left(a_{n+1}\right)=\frac{1}{f\left(-2-a_{n}\right)}$, we get
$$f\left(a_{n+1}\right) f\left(-2-a_{n}\right)=1$$

Thus $f\left(a_{n+1}-a_{n}-2\right)=f(0), a_{n+1}-a_{n}-2=0$.
$$a_{n+1}-a_{n}=2(n \in \mathbf{N})$$

Therefore, $\left\{a_{n}\right\}$ is an arithmetic sequence with the first term 1 and common difference 2, hence $a_{n}=2 n-1, a_{2003}=$ 4005.
(2) From $\left[1+\frac{1}{a_{1}}\right]\left[1+\frac{1}{a_{2}}\right] \cdots\left[1+\frac{1}{a_{n}}\right] \geqslant k \sqrt{2 n+1}$ always holds, we know
$$k \leqslant \frac{\left[1+\frac{1}{a_{1}}\right]\left[1+\frac{1}{a_{2}}\right] \cdots\left[1+\frac{1}{a_{n}}\right]}{\sqrt{2 n+1}}$$

always holds.
Let $F(n)=\frac{\left[1+\frac{1}{a_{1}}\right]\left[1+\frac{1}{a_{2}}\right] \cdots\left[1+\frac{1}{a_{n}}\right]}{\sqrt{2 n+1}}$, then
and
$$F(n+1)=\frac{\left[1+\frac{1}{a_{1}}\right]\left[1+\frac{1}{a_{2}}\right] \cdots\left[1+\frac{1}{a_{n+1}}\right]}{\sqrt{2 n+3}}$$

Also, $\frac{F(n+1)}{F(n)}=\frac{2(n+1)}{\sqrt{4(n+1)^{2}-1}}>1$, i.e., $F(n+1)>F(n)$, hence $F(n)$ is a monotonically increasing function of $n$, $F(n) \geqslant F(1)=\frac{2}{3} \sqrt{3}$.