Example 1: At each vertex of a regular 2009-gon, a non-negative integer not exceeding 100 is placed. Adding 1 to the numbers at two adjacent vertices is called an operation on these two adjacent vertices. For any given two adjacent vertices, the operation can be performed at most $k$ times. Find the minimum value of $k$ such that it is always possible to make all the numbers at the vertices equal. ${ }^{[1]}$
(35th Russian Mathematical Olympiad)
Solve $k_{\min }=100400$.
Let the numbers at vertices $A_{1}, A_{2}, \cdots, A_{2009}$ be $a_{1}, a_{2}, \cdots, a_{2009}$ respectively.
First, take $a_{2}=a_{4}=a_{2008}=100, a_{1}=a_{3}=a_{2009}=0$.
For each operation, assign
$$
S=\left(a_{2}-a_{3}\right)+\left(a_{4}-a_{5}\right)+\cdots+\left(a_{2008}-a_{2009}\right) \text {. }
$$

If the adjacent vertex pair $\left(A_{1}, A_{2}\right)$ is operated once, then $S$ increases by 1;

If the adjacent vertex pair $\left(A_{1}, A_{2009}\right)$ is operated once, then $S$ decreases by 1;

If other adjacent vertex pairs $\left(A_{i}, A_{i+1}\right)(i=2,3 \cdots, 2008)$ are operated once, then $S$ remains unchanged.
Initially, $S=100400$; finally, $S=0$.
Thus, the adjacent vertex pair $\left(A_{1}, A_{2009}\right)$ must be operated at least 100400 times.
Therefore, $k \geqslant 100400$.
Next, we prove that for any initial state $\left(a_{1}, a_{2}, \cdots, a_{2009}\right)$ $\left(a_{j} \in \mathbf{N}, a_{j} \leqslant 100\right)$, all numbers can be made equal through at most 100400 operations.
In fact, for the adjacent vertex pair $\left(A_{i}, A_{i+1}\right)$, operate
$$
S_{i}=a_{i+2}+a_{i+4}+\cdots+a_{i+2008}
$$

times (with the convention $a_{j+2009}=a_{j}$), then $a_{i}$ will eventually become
$$
a_{i}+S_{i-1}+S_{i}=a_{1}+a_{2}+\cdots+a_{2009} \text {. }
$$

Thus, all numbers become equal, and the number of operations on the adjacent vertex pair $\left(A_{i}, A_{i+1}\right)$
$$
\begin{array}{l}
S_{i}=a_{i+2}+a_{i+4}+\cdots+a_{2008} \\
\leqslant 100 \times 1004=100400,
\end{array}
$$

which meets the problem's requirements.