3 Given $a_{1}, a_{2}, \cdots, a_{11}$ are 11 distinct positive integers with a total sum 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 many? (Provided by Liu Zhiming)
Note that a permutation $x_{1}, x_{2}, \cdots, x_{n}$ of $1,2, \cdots, n$ is called an even permutation if $\prod_{i>j}\left(x_{i}-x_{j}\right)$ is positive; otherwise, it is called an odd permutation.
Solution: The number of good operation groups is more, by $\prod_{i=1}^{11} a_{i}$.
We introduce the general notation: If the blackboard writes the numbers $1,2, \cdots, n$, an operation group is defined as $l$ consecutive operations: the $i$-th operation can select any one of 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 $1,2, \cdots, n$, then this operation group is called good (suboptimal), and the difference between the number of good and suboptimal operation groups is denoted as $f\left(b_{1}, b_{2}, \cdots, b_{l} ; n\right)$. We will first 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 good (suboptimal) operation groups remains unchanged, and thus $f$ remains unchanged.

Second, we only need to calculate the number of good (suboptimal) operation groups such that 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 good and 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 good and suboptimal operation groups is the same. 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, good operation groups can be paired one-to-one with suboptimal operation groups through the above changes, so in the operation groups without property $P$, the number of good and suboptimal operation groups is the same.
Now we use mathematical induction on $m$ to prove the following result:
$a_{1}, a_{2}, \cdots, a_{m}$ are $m$ distinct positive integers with a total sum less than $n$, then
$$
f\left(a_{1}, a_{2}, \cdots, a_{m},-a_{1},-a_{2}, \cdots,-a_{m} ; n\right)=\prod_{j=1}^{m} a_{j} .
$$

When $m=1$, consider the good 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 $1,2, \cdots, n$), so there are $a_{1}$ good operation groups and 0 suboptimal operation groups, hence (1) holds.

Assume the proposition holds for $m-1$. Consider the proposition for $m$. Without loss of generality, assume $a_{1}<$ $a_{2}<\cdots<a_{n}$. According to the previous discussion,
$$
\begin{aligned}
& f\left(a_{1}, a_{2}, \cdots, a_{m},-a_{1},-a_{2}, \cdots,-a_{m} ; n\right) \\
= & f\left(a_{1},-a_{2},-a_{3}, \cdots,-a_{m}, a_{2}, a_{3}, \cdots, a_{m},-a_{1} ; n\right),
\end{aligned}
$$

For the good 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 $1+2+\cdots+\left(n-a_{1}\right)$, and the $n$ numbers on the blackboard after the operation group will not be a permutation of $1,2, \cdots, n$, 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 $1,2, \cdots, n-a_{1}$, so that the total $2 m$ steps can result in an even (odd) permutation of $1,2, \cdots, n$. So, each good (suboptimal) operation group with property $P$ for the $n-a_{1}$ numbers can correspond to $a_{1}$ original good (suboptimal) operation groups with property $P$, thus
$$
\begin{aligned}
& f\left(a_{1},-a_{2},-a_{3}, \cdots,-a_{m}, a_{2}, a_{3}, \cdots, a_{m},-a_{1} ; n\right) \\
= & a_{1} \cdot f\left(-a_{2},-a_{3}, \cdots,-a_{m}, a_{2}, a_{3}, \cdots, a_{m} ; n-a_{1}\right) .
\end{aligned}
$$

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

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