10. 4. In a regular 2009-gon, a non-negative integer not exceeding 100 is placed at each vertex. 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.
10. 4. $k_{\min }=100400$.

Let the numbers at each vertex be $a_{1}, a_{2}, \cdots, a_{2009}$. Let $N=100400$.
$$
\begin{array}{l}
\text { (1) Let } a_{2}=a_{4}=\cdots=a_{2008}=100, \\
a_{1}=a_{3}=\cdots=a_{2008}=0, \\
S=\left(a_{2}-a_{3}\right)+\left(a_{4}-a_{5}\right)+\cdots+\left(a_{2008}-a_{2009}\right) .
\end{array}
$$

For adjacent vertices $\left(a_{1}, a_{2}\right)$, $S$ increases by 1 after one operation; for adjacent vertices $\left(a_{1}, a_{2009}\right)$, $S$ decreases by 1 after one operation; for other adjacent vertices, $S$ remains unchanged.
Initially, $S=N$, and finally, $S=0$.
Therefore, at least $N$ operations must be performed on adjacent vertices $\left(a_{1}, a_{2009}\right)$. Thus, $k \geqslant N$.
(2) Below, we prove that all numbers can be made equal through at most $N$ operations.
For adjacent vertices $\left(a_{i}, a_{i+1}\right)$, perform the operation
$$
s_{i}=a_{i+2}+a_{i+4}+\cdots+a_{i+2 \text { 008 }} \text { (times). }
$$

Then $a_{i}$ will eventually become
$$
a_{i}+s_{i-1}+s_{i}=a_{1}+a_{2}+\cdots+a_{2000} .
$$

And the number of operations on adjacent vertices $\left(a_{i}, a_{i+1}\right)$
$$
s_{i} \leqslant 1004 \times 100=N .
$$