On a circle there are 2009 nonnegative integers not greater than 100. If two numbers sit next to each other, we can increase both of them by 1. We can do this at most $ k$ times. What is the minimum $ k$ so that we can make all the numbers on the circle equal?
1. **Define the problem and variables:**
   We are given a circle with \( n = 2009 \) nonnegative integers, each not greater than 100. We can increase any two adjacent numbers by 1, and we need to determine the minimum number \( k \) of such operations required to make all the numbers equal.

2. **Express the problem in terms of variables:**
   Let \( a_1, a_2, \ldots, a_{2m+1} \) be the numbers on the circle, where \( n = 2m + 1 \). Let \( x_i \geq 0 \) denote the number of times we increase the numbers at positions \( i \) and \( i+1 \) by 1.

3. **Set up the system of equations:**
   The numbers are equal in the end if and only if:
   \[
   a_i + x_{i-1} + x_i = a_{i+1} + x_i + x_{i+1} \quad \text{for all } i
   \]
   Simplifying, we get:
   \[
   x_{i-1} - x_{i+1} = a_{i+1} - a_i
   \]
   This gives us a system of linear equations:
   \[
   \begin{align*}
   x_2 - x_{2m+1} &= a_1 - a_2 \\
   x_4 - x_2 &= a_3 - a_4 \\
   &\vdots \\
   x_{2m-2} - x_{2m-4} &= a_{2m-3} - a_{2m-2} \\
   x_{2m} - x_{2m-2} &= a_{2m-1} - a_{2m} \\
   x_{2m-1} - x_{2m+1} &= a_{2m+1} - a_{2m} \\
   x_{2m-3} - x_{2m-1} &= a_{2m-1} - a_{2m-2} \\
   &\vdots \\
   x_3 - x_5 &= a_5 - a_4 \\
   x_1 - x_3 &= a_3 - a_2
   \end{align*}
   \]
   The condition \( x_{2m} - x_1 = a_1 - a_{2m+1} \) is redundant.

4. **Find the minimum \( k \):**
   We need to find the smallest \( k_m \) such that for any \( (a_i)_{i=1}^{2m+1} \in \{0, 1, \ldots, 100\}^{2m+1} \), there exists a solution \( (x_1, \ldots, x_{2m+1}) \) to the system such that \( x_1 + \cdots + x_{2m+1} \leq k_m \).

5. **Show that \( k_m \geq 100m(m+1) \):**
   Consider the sequence \( (a_1, a_2, \ldots, a_{2m+1}) = (100, 0, 100, 0, \ldots, 100, 0, 100) \). We must have:
   \[
   x_{2j} = x_{2m+1-j} = x_{2m+1} + 100j \quad \text{for all } 1 \leq j \leq m
   \]
   Therefore:
   \[
   x_1 + \cdots + x_{2m+1} = (2m+1)x_{2m+1} + 100(1 + 2 + \cdots + m) + 100(1 + 2 + \cdots + m) \geq 100m(m+1)
   \]

6. **Show that \( k_m = 100m(m+1) \) suffices:**
   For any \( (a_1, \ldots, a_{2m+1}) \in \{0, 1, \ldots, 100\}^{2m+1} \), there exists a solution with \( \min\{x_1, \ldots, x_{2m+1}\} = 0 \). Suppose \( x_{2m+1} = 0 \). Then \( |a_i - a_{i+1}| \leq 100 \) for all \( i \) forces \( x_{2j}, x_{2m+1-2j} \leq 100j \) for \( 1 \leq j \leq m \). Thus:
   \[
   x_1 + \cdots + x_{2m+1} \leq 0 + 100(1 + 2 + \cdots + m) + 100(1 + 2 + \cdots + m) = 100m(m+1)
   \]
   Hence, \( k_m = 100m(m+1) \) suffices.

\(\blacksquare\)

The final answer is \( \boxed{ 100m(m+1) } \).