Three, given that $a_{1}, a_{2}, \cdots, a_{11}$ are 11 distinct positive integers, and their total sum is less than 2007. On the blackboard, the numbers $1, 2, \cdots, 2007$ are written in sequence. Define a sequence of 22 consecutive operations as an operation group: the $i$-th operation can select any number from the existing numbers on the blackboard, and when $1 \leqslant i \leqslant 11$, add $a_{i}$; when $12 \leqslant i \leqslant 22$, subtract $a_{i-11}$. If the final result is an even permutation of $1, 2, \cdots, 2007$, then this operation group is called excellent; if the final result is an odd permutation of $1, 2, \cdots, 2007$, then this operation group is called suboptimal. Question: Which is more, excellent operation groups or suboptimal operation groups, and by how much?

Note: A permutation $x_{1}, x_{2}, \cdots, x_{n}$ of $1, 2, \cdots, n$ is called an even permutation if $\sum_{i>j}\left(x_{i}-x_{j}\right)$ is positive; otherwise, it is called an odd permutation.
（Liu Zhige, problem contributor）
3. The number of excellent operation groups is more by $\prod_{i=1}^{11} a_{i}$.

We introduce a general notation: If the blackboard has the numbers 1, 2, ..., $n$, an operation group is defined as $l$ consecutive operations: the $i$-th operation can select any number from the existing numbers on the blackboard and add $b_{i}$, where $b_{i} \in \mathbf{Z} (1 \leqslant i \leqslant l)$. If the final result is an even (odd) permutation of 1, 2, ..., $n$, then this operation is called excellent (suboptimal). The difference between the number of excellent operation groups and the number of suboptimal operation groups is denoted as $f\left(b_{1}, b_{2}, \cdots, b_{l} ; n\right)$.
We now discuss the properties of $f$.
First, for any $1 \leqslant i, j \leqslant l$, swapping the values of $b_{i}$ and $b_{j}$ does not affect $f$. In fact, it is only necessary to swap the $i$-th and $j$-th operations in the operation group; in other words, the $i$-th operation performs the original $j$-th operation, adding $b_{j}$ to the number selected for the original $j$-th operation, and the $j$-th operation performs the original $i$-th operation. The result of the operation group remains unchanged after the swap, so the number of excellent (suboptimal) operation groups remains unchanged, and thus $f$ remains unchanged.

Second, it is only necessary to calculate the number of such excellent (suboptimal) operation groups: no two numbers on the blackboard are the same after each step of the operation, such operation groups are called operation groups with property $P$. It can be proven that the difference between the number of excellent operation groups with property $P$ and the number of suboptimal operation groups with property $P$ is also equal to $f\left(b_{1}, b_{2}, \cdots, b_{l} ; n\right)$.

In fact, it is only necessary to prove that in the operation groups without property $P$, the number of excellent operation groups is the same as the number of suboptimal operation groups. If an operation group first causes two equal numbers on the blackboard at the $i$-th step, for example, the $p$-th number and the $q$-th number are equal $(1 \leqslant p < q \leqslant n)$, then the following changes can be made to the last $l-i$ steps of the operation group: the operation on the $p$-th number is changed to the operation on the $q$-th number, and the operation on the $q$-th number is changed to the operation on the $p$-th number. The final result of this new operation group will be the original operation group's result with the $p$-th and $q$-th numbers swapped. It is not difficult to see that swapping any two numbers in a permutation will change the parity of the permutation. Therefore, excellent operation groups can be paired one-to-one with suboptimal operation groups through the above changes. Thus, in the operation groups without property $P$, the number of excellent and suboptimal operation groups is the same.
Now we prove by induction on $m$:
If $a_{1}, a_{2}, \cdots, a_{m}$ are $m$ distinct positive integers with a total sum less than $n$, then
$$
\begin{array}{l}
f\left(a_{1}, a_{2}, \cdots, a_{m},-a_{1},-a_{2}, \cdots,-a_{m} ; n\right) \\
=\prod_{j=1}^{m} a_{j} .
\end{array}
$$

When $m=1$, consider the excellent and suboptimal operation groups with property $P$. They must select a number from the last $a_{1}$ numbers and add $a_{1}$, and then add $-a_{1}$ to this number (otherwise, it will not result in a permutation of 1, 2, ..., $n$). Therefore, there are $a_{1}$ excellent operation groups and 0 suboptimal operation groups. Hence, equation (1) holds.
Assume that the proposition holds for $m-1$. Consider the proposition for $m$.
Without loss of generality, assume $a_{1} < a_{2} < \cdots < a_{m}$. According to the previous discussion,
$$
\begin{array}{l}
f\left(a_{1}, a_{2}, \cdots, a_{m},-a_{1},-a_{2}, \cdots,-a_{m} ; n\right) \\
=f\left(a_{1},-a_{2},-a_{3}, \cdots,-a_{m}, a_{2}, a_{3}, \cdots, a_{m},-a_{1} ; n\right),
\end{array}
$$

For the excellent and suboptimal operation groups with property $P$, the first step can select a number from the last $a_{1}$ numbers and add $a_{1}$, and the second step can only operate on the first $a_{2}$ numbers, the third step can only operate on the first $a_{2}+a_{3}$ numbers, ..., the $m$-th step can only operate on the first $a_{2}+\cdots+a_{m}<n-a_{1}$ numbers, and the $(m+1) \sim 2 m-1$ steps can only operate on the first $n-a_{1}$ numbers (otherwise, the sum of the first $n-a_{1}$ numbers will be less than $1+2+\cdots+(n-a_{1})$, and the $n$ numbers on the blackboard after the operation group will not be a permutation of 1, 2, ..., $n$), and the $2 m$-th step can only operate on the number selected in the first step.

Therefore, the $2 \sim 2 m-2$ steps must operate on the first $n-a_{1}$ numbers, and the result must be an even (odd) permutation of 1, 2, ..., $n-a_{1}$, so that the result of the total $2 m$ steps is an even (odd) permutation of 1, 2, ..., $n$. Thus, each excellent (suboptimal) operation group with property $P$ for the $n-a_{1}$ numbers can correspond to $a_{1}$ original excellent (suboptimal) operation groups with property $P$. Therefore,
$$
\begin{array}{l}
f\left(a_{1},-a_{2},-a_{3}, \cdots,-a_{m}, a_{2}, a_{3}, \cdots, a_{m},-a_{1} ; n\right) \\
=a_{1} f\left(-a_{2},-a_{3}, \cdots,-a_{m}, a_{2}, a_{3}, \cdots, a_{m} ; n-a_{1}\right) .
\end{array}
$$

According to the induction hypothesis
$$
\begin{array}{l}
f\left(-a_{2},-a_{3}, \cdots,-a_{m}, a_{2}, a_{3}, \cdots, a_{m} ; n-a_{1}\right) \\
=f\left(a_{2}, a_{3}, \cdots, a_{m},-a_{2},-a_{3}, \cdots,-a_{m} ; n-a_{1}\right) \\
=\prod_{j=2}^{\infty} a_{j} .
\end{array}
$$

Thus, equation (1) also holds for $m$.
Therefore, by induction, equation (1) is proven.
Taking $n=2007, m=11$ in equation (1), we get the answer
$$
\text { answer } \prod_{j=1}^{11} a_{j} \text {. }
$$