18. (1) When $x_{1}=\frac{1}{3}, x_{n+1}=x_{n}^{2}+x_{n}$, in which interval of two consecutive integers does the expression $\frac{1}{1+x_{1}}+\frac{1}{1+x_{2}}+\cdots+\frac{1}{1+x_{2001}}+$ $\frac{1}{1+x_{2002}}$ lie? (2002 2003 Finnish Mathematical Olympiad Problem)
18. (1) Since $x_{n}^{2}+x_{n}=x_{n}\left(\overline{x_{n}}+1\right)$, and
$$\frac{1}{x_{n}\left(x_{n}+1\right)}=\frac{1}{x_{n}}-\frac{1}{x_{n}+1}, \frac{1}{x_{n}+1}=\frac{1}{x_{n}}-\frac{F}{x_{n+1}}$$

Therefore,
$$\sum_{n=1}^{2002} \frac{1}{x_{n}+1}=\frac{1}{x_{1}}-\frac{1}{x_{2003}}=3-\frac{1}{x_{2003}}$$

By specifically calculating $x_{2}, x_{3}, x_{4}$, we have $x_{2}=\frac{1}{3}\left(\frac{1}{3}+1\right)=\frac{4}{9}, x_{3}=\frac{4}{9}\left(\frac{4}{9}+1\right)=\frac{52}{81}, x_{4}=\frac{52}{81}$ $\left(\frac{52}{81}+1\right)=\frac{6916}{6561}>1$, since the sequence $\left\{x_{n}\right\}$ is monotonically increasing, so, $0<\frac{1}{x_{2003}}<1$, therefore, the sum of the algebraic expression is between 2 and 3.