Let $a_1, a_2, \ldots , a_{11}$ be 11 pairwise distinct positive integer with sum less than 2007. Let S be the sequence of  $1,2, \ldots ,2007$. Define an [b]operation[/b] to be 22 consecutive applications of the following steps on the sequence $S$: on $i$-th step, choose a number from the sequense $S$ at random, say $x$. If $1 \leq i \leq 11$, replace $x$ with $x+a_i$ ; if $12 \leq i \leq 22$, replace $x$ with $x-a_{i-11}$ . If the result of [b]operation[/b] on the sequence $S$ is an odd permutation of $\{1, 2, \ldots , 2007\}$, it is an [b]odd operation[/b]; if the result of [b]operation[/b] on the sequence $S$ is an even permutation of $\{1, 2, \ldots , 2007\}$, it is an [b]even operation[/b]. Which is larger, the number of odd operation or the number of even permutation? And by how many?

Here $\{x_1, x_2, \ldots , x_{2007}\}$ is an even permutation of $\{1, 2, \ldots ,2007\}$ if the product $\prod_{i > j} (x_i - x_j)$ is positive, and an odd one otherwise.
To solve this problem, we need to determine whether the number of odd operations is larger than the number of even operations, and by how many. We will use the properties of permutations and the given operations to derive the solution.

1. **Define the Operations:**
   - We start with the sequence \( S = \{1, 2, \ldots, 2007\} \).
   - We perform 22 consecutive steps on \( S \):
     - For \( 1 \leq i \leq 11 \), we choose a number \( x \) from \( S \) and replace it with \( x + a_i \).
     - For \( 12 \leq i \leq 22 \), we choose a number \( x \) from \( S \) and replace it with \( x - a_{i-11} \).

2. **Understand the Permutations:**
   - A permutation of \( \{1, 2, \ldots, 2007\} \) is even if the product \( \prod_{i > j} (x_i - x_j) \) is positive, and odd if it is negative.
   - The operations described will result in a permutation of the original sequence \( S \).

3. **Observation and Proof:**
   - Let \( b_j = a_j \) for \( 1 \leq j \leq 11 \) and \( b_j = -a_{j-11} \) for \( 12 \leq j \leq 22 \).
   - If two elements become equal after adding \( b_1, \ldots, b_l \), the total contribution of \( b_1, \ldots, b_l \) at these fixed spots is 0.
   - Suppose after some steps, \( s_i = s_j \). The sums of the added values to \( s_i \) and \( s_j \) must be equal, implying that the sequence is not a permutation of \( \{1, \ldots, 2007\} \).

4. **Inductive Argument:**
   - If \( a_i \) is added to \( s_j \) and \( s_j + a_i \leq 2007 \), we can assume the moves are ordered such that \( a_i \) is added to \( s_j \) first.
   - This implies that \( a_1 \) cannot be added to \( 1, \ldots, 2007 - a_1 \), then \( a_2 \) cannot be added to \( 1, \ldots, 2007 - a_2 \), and so on.
   - After a number gets added, it must be subtracted because otherwise, it would exceed 2007.

5. **Reordering Moves:**
   - Reorder the moves such that \( a_1 \) is added first, and the second move is subtracting the number that \( a_1 \) was added to.
   - If the second move is \( -a_1 \), the two moves cancel out.
   - If the second move is \( -a_2 \), the new number is \( x + a_1 - a_2 \). Since \( x + a_1 > 2007 \), it is positive, and since \( a_1 < a_2 \), the new number is less than \( x \), making two numbers equal, resulting in a contribution of 0.

6. **Counting the Operations:**
   - There are \( a_1 \) places to perform this operation, namely \( 2007 - a_1 + 1, \ldots, 2007 \).
   - By induction, adding and subtracting \( a_2, a_3, \ldots, a_k \) gives a total contribution of \( \prod_{j=2}^k a_j \) ways, leading to \( \prod_{j=1}^k a_j \) ways.

7. **Conclusion:**
   - Since the base case can be easily verified, the conclusion follows that the number of odd operations and even operations are equal.

The final answer is \(\boxed{0}\) (the difference between the number of odd and even operations).