Given an infinite sequence $\left\{a_{n}\right\}$ where all terms are positive integers, and the sum of any consecutive terms is not equal to 100. Find the minimum value of $\max \left\{a_{n}, n \in \mathbf{N}\right\}$.
Solution: The minimum value of $\max \left\{a_{n}, n \in \mathbf{N}\right\}$ is 3.
First, we prove: If all terms of the sequence $\left\{a_{n}\right\}$ are 1 or 2, then there must exist a continuous subsequence whose sum equals 100.
Consider the first 100 terms of the sequence $\left\{a_{n}\right\}$, and let
$$
S_{0}=0, S_{1}=a_{1}, \cdots, S_{k}=\sum_{i=1}^{k} a_{i}, k \leqslant 100 .
$$

By the pigeonhole principle, there exist $0 \leqslant m<n \leqslant 100$ such that $S_{n}-S_{m}=100 t$, i.e.,
$$
\sum_{i=m+1}^{n} a_{i}=100 t .
$$

Since $a_{i}=1$ or 2, then $t=1$ or 2.
If $t=1$, the conclusion holds.
If $t=2$, by $a_{i} \leqslant 2, n \leqslant 100$, we get
$$
200 \geqslant \sum_{i=1}^{100} a_{i} \geqslant \sum_{i=m+1}^{n} a_{i}=200 .
$$

Then $a_{i}=2, i=1,2, \cdots, 100$. Hence $\sum_{i=1}^{50} a_{i}=100$.
In this case, the conclusion still holds. Therefore, $\max \left\{a_{n}, n \in \mathbf{N}\right\} \geqslant 3$.
Next, we construct an infinite periodic sequence such that the largest term is 3. The first 100 terms of the sequence $\left\{a_{n}\right\}$ are
$$
1,2, \cdots, 2,3,2, \cdots, 2, \cdots,
$$

where $a_{100 k+1}=1, a_{100 k+51}=3, k=0,1, \cdots$, and the other terms $a_{i}=2$.

It is easy to verify that the sum of any continuous subsequence of this sequence does not equal 100.
In summary, the minimum value of $\max \left\{a_{n}, n \in \mathbf{N}\right\}$ is 3.
(Zhang Yanwei, Suqian Education Bureau, Jiangsu Province, 223800)