A basket is called "[i]Stuff Basket[/i]" if it includes $10$ kilograms of rice and $30$ number of eggs. A market is to distribute $100$ Stuff Baskets. We know that there is totally $1000$ kilograms of rice and $3000$ number of eggs in the baskets, but some of market's baskets include either more or less amount of rice or eggs. In each step, market workers can select two baskets and move an arbitrary amount of rice or eggs between selected baskets. Starting from an arbitrary situation, what's the minimum number of steps that workers provide $100$ Stuff Baskets?
1. **Define Variables:**
   Let \( a_i \) be the amount of rice in basket \( i \) and \( b_i \) be the number of eggs in basket \( i \). We need to ensure that each basket has exactly 10 kg of rice and 30 eggs.

2. **Initial Conditions:**
   We are given that the total amount of rice is 1000 kg and the total number of eggs is 3000. Therefore, the sum of rice in all baskets is:
   \[
   \sum_{i=1}^{100} a_i = 1000
   \]
   and the sum of eggs in all baskets is:
   \[
   \sum_{i=1}^{100} b_i = 3000
   \]

3. **Existence of Sufficient Resources:**
   We need to show that there exist two baskets \( i \) and \( j \) such that \( a_i + a_j \geq 10 \) and \( b_i + b_j \geq 30 \). 

4. **Proof for Eggs:**
   Suppose \( b_m \) is the maximum number of eggs in any basket. If \( b_m < 30 \), then:
   \[
   b_1 + b_2 + \cdots + b_{100} < 100b_m < 3000
   \]
   which contradicts the given total number of eggs. Therefore, \( b_m \geq 30 \).

5. **Proof for Rice:**
   If \( a_m \geq 10 \), then we are done. Otherwise, suppose \( a_m < 10 \). We need to show there exists a basket \( j \) such that \( a_m + a_j > 10 \). Assume for contradiction that \( a_i + a_m \leq 10 \) for all \( i \neq m \). Then:
   \[
   (a_1 + a_m) + (a_2 + a_m) + \cdots + (a_{m-1} + a_m) + (a_{m+1} + a_m) + \cdots + (a_{100} + a_m) \leq 990
   \]
   Simplifying, we get:
   \[
   (a_1 + a_2 + \cdots + a_{100}) + 99a_m \leq 990
   \]
   Given \( a_1 + a_2 + \cdots + a_{100} = 1000 \), we have:
   \[
   1000 + 99a_m \leq 990
   \]
   which is a contradiction since \( 1000 + 99a_m \geq 1000 \). Therefore, there must exist a basket \( j \) such that \( a_m + a_j > 10 \).

6. **Constructing Stuff Baskets:**
   Using the above results, we can always find two baskets to combine resources and create a Stuff Basket. Each step involves selecting two baskets and redistributing their contents to form one Stuff Basket.

7. **Minimum Steps:**
   Since each step creates one Stuff Basket and reduces the number of baskets needing adjustment by one, we need \( 99 \) steps to create \( 100 \) Stuff Baskets from an arbitrary initial configuration.

\[
\boxed{99}
\]