I Let $X$ be a subset of $\mathbf{N}^{*}$, the smallest element of $X$ is 1, and the largest element is 100. For any number in $X$ that is greater than 1, it can be expressed as the sum of two numbers (which can be the same) in $X$. Find the minimum value of $|X|$.
1. $|X|_{\min }=9$. Let $X=\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$, and $1=a_{1}<a_{2}<\cdots<a_{n}=100$. For any $a_{k} (2 \leq k \leq n-1)$, there exist $a_{i}, a_{j}$ such that $a_{k}=a_{i}+a_{j}$, where it must be that $a_{i}<a_{k}, a_{j}<a_{k}$, i.e., $a_{i} \leqslant a_{k-1}, a_{j} \leqslant a_{k-1}$. Therefore, $a_{k}=a_{i}+a_{j} \leqslant a_{k-1}+a_{k-1}=2 a_{k-1}$. So $100=a_{n} \leqslant 2 a_{n-1} \leqslant 4 a_{n-2} \leqslant \cdots \leqslant 2^{n-1} a_{1}=2^{n-1}$, hence $n \geqslant 8$. If $n=8$, then $1=a_{1}<a_{2}<\cdots<a_{8}=100$. It is easy to see that $a_{2}=2=2^{2-1}$, otherwise $a_{2}$ cannot be expressed as the sum of two numbers in $X$. Now, note that $a_{8}=100$ is not a power of 2, so we can assume that $X$ contains $a_{k}=2^{k-1} (k=1,2, \cdots, t-1), a_{t} \neq 2^{t-1}$. Since $a_{t}$ can be expressed as the sum of two numbers in $X$, there exist $a_{i}, a_{j} (i, j < t)$ such that $a_{t}=a_{i}+a_{j}$. If $a_{i}=a_{j}$, then $a_{t}=2 a_{i}=2 \times 2^{i-1}=2^{i}$. If $i=t-1$, then it contradicts $a_{t} \neq 2^{t-1}$. So $i < t-1$, and thus $a_{t}=2 a_{i}=2^{i}=a_{i+1}$, which contradicts the distinctness of the numbers in $X$. Therefore, $a_{i} \neq a_{j}$. So $a_{j}<2^{t-2}, a_{i}<2^{t-3}$, and thus $a_{t}=a_{i}+a_{j}<2^{t-2}+2^{t-3}$. Therefore, $100=a_{8}=2^{8-t} a_{t}<2^{8-t}(2^{t-2}+2^{t-3})=2^{6}+2^{5}=96$, which is a contradiction. Therefore, $n \neq 8$, and thus $n \geqslant 9$.