China Mathematical Olympiad

The answer is: The "good" groups is more than the "second good" groups by ${\prod }_{i=1}^{11}{a}_i$.

More generally, we write numbers $1,2,\dots ,n$ in order on the blackboard, and define a group of $l$ ordered operations: The $i$-th operation is to take any number on the blackboard, and then add $b_i$ ($b_i\in \mathbb{Z},1\le i\le l$) to it. If the final result after such a group of operations is an even/odd permutation of $1,2,\dots ,n$, then we call it a "good"/"second good" group. And the difference between the number of "good" groups and that of "second good" groups is defined as $$f(b_1,b_2,\cdots ,b_l;n).$$ Now let us study the property of $f$.

Firstly, interchanging $b_i$ and $b_j$ for any $1\le i,j\le l$ will not affect the value of $f$. As a matter of fact, it only results in the exchange of the $i$th and $j$th operations in a group, and will not affect the final result after the group's operations. So the value of $f$ remains the same.

Secondly, we only need to count the number of "good"/"second good" groups with property $P$ — a property attributed to any operation group which keeps the numbers on the blackboard distinctive from one another after each operation. We can prove that the difference between the numbers of "good" and "second good" groups with property $P$ is also equal to $f$.

In fact, we only need to prove that the numbers of "good" and "second good" groups without property $P$ are the same. Suppose the $i$th operation of a "good"/"second good" group without property $P$ results in the equal between the $p$th and $q$th number on the blackboard ($1\le p<q\le n$). We change the following $l-i$ operations in this way: operations on the $p$th number are changed to operations on the $q$th number, and vice versa. It is easy to verify that the resulted permutation on the blackboard of new operation group would be a $(p,q)$ transposition of the permutation of the original operation group. Then the parities of the two permutations are in opposite signs. And that means the numbers of "good" and "second good" groups without property $P$ are the same.

Now, let $a_1,a_2,\cdots ,a_m$ be $m$ distinct positive integers with their sum less than $n$. We prove by the principle of mathematical induction that $$f(a_1,a_2,\dots ,a_m,-a_1,-a_2,\dots ,-a_m;n)=\prod _{i=1}^m{a}_i.(1)$$ If $m=1$, consider a "good"/"second good" group with property $P$. It must be in such a way: The first operation is to take a number from $n-a_1+1,n-a_2+2,\dots ,n$ on the blackboard, and add $a_1$ to it; next operation is to add $-a_1$ again to it. So the number of "good" groups is $a_1$, while that of "second good" groups is $0$. Therefore $(1)$ holds.

Assume that $(1)$ holds for $m-1$. We now consider case $m$. According to what discussed above, we may assume that $a_1<a_2<\cdots <a_n$, and $$\begin{array}{rl}& f(a_1,a_2,\dots ,a_m,-a_1,-a_2,\dots ,-a_m;n)\\ & =f(a_1,-a_2,-a_3,\dots ,-a_m,a_2,a_3,\dots ,a_m,-a_1;n).\end{array}$$ For a group with property $P$, the first operation must be done on the last $a_1$ numbers on the blackboard; the second operation be done on the first $a_2$ numbers; the third operation done on the first $a_2+a_3$ numbers; ... the $m$th operation done on the first $a_2+\cdots +a_m<n-a_1$ numbers. And the $m+1\sim 2m-1$ th operations will also be done on the first $n-a_1$ numbers. Otherwise the sum of the first $n-a_1$ numbers will be less than $1+2+\cdots +(n-a_1)$, a contradiction to property $P$.

Therefore, the $2\sim 2m-2$ operations must be done on the first $n-a_1$ numbers, and the result must be an even/odd permutation of $1,2,\dots ,(n-a_1)$, which corresponds to each one of $a_1$ even/odd permutations of $1,2,\dots ,n$ derived from original operation groups. therefore $$\begin{array}{rl}& f(a_1,-a_2,-a_3,\dots ,-a_m,a_2,a_3,\dots ,a_m,-a_1;n)\\ & =a_{1}f(-a_2,-a_3,\dots ,-a_m,a_2,a_3,\dots ,a_m;n-a_1).\end{array}$$ By induction we have $$\begin{array}{rl}& f(-a_2,-a_3,\cdots ,-a_m,a_2,a_3,\cdots ,a_m;n-a_1)\\ & =f(a_2,a_3,\cdots ,a_m,-a_2,-a_3,\cdots ,-a_m;n-a_1)\\ & =\prod _{j=2}^m{a}_j.\end{array}$$ That means $(1)$ holds for $m$.

Now take $n=2007$ and $m=11$ in $(1)$. Thus arriving at the value of ${\prod }_{j=1}^{11}{a}_j$.