3. 2. 14 * The sequence $\left\{x_{n}\right\}$ is defined as follows: $x_{1}=\frac{1}{2}, x_{n+1}=x_{n}^{2}+x_{n}$, find the integer part of the following sum: $\frac{1}{1+x_{1}}+\frac{1}{1+x_{2}}+\cdots+\frac{1}{1+x_{2 \times 00}}$.
From the given conditions, we know that for any positive integer $n$, we have $x_{n}>0$. Furthermore, it is also known that when $n \geqslant 3$, $x_{n}>1$. From the problem statement, we know that $\frac{1}{x_{n+1}}=\frac{1}{x_{n}\left(1+x_{n}\right)}=\frac{1}{x_{n}}-\frac{1}{1+x_{n}}$, which means $\frac{1}{1+x_{n}}=\frac{1}{x_{n}}-\frac{1}{x_{n+1}}$. Therefore, $\sum_{k=1}^{2000} \frac{1}{1+x_{k}}=\sum_{k=1}^{2000}\left(\frac{1}{x_{k}}-\frac{1}{x_{k+1}}\right)=\frac{1}{x_{1}}-\frac{1}{x_{2001}}=2-\frac{1}{x_{2001}}$.
From the previous discussion, we know that $0<\frac{1}{x_{2001}}<1$. Therefore, the integer part of the required sum is 1.