Let $n$ be a positive integer. Let $m$ be the smallest positive integer with
$m > n - m$. Since $n > n - n = 0$, $m \le n$. Let $N = \{1, \ldots, n\}$
be the set of positive integers from $1$ to $n$. Let $M \subset N$,
be $M = \{1, \ldots, m\}$.
Let $P_n$ be the set of of permutations of $N$.
Let $C_m$ be the set of cyclic permutations of $M$, there are $m$
cyclic permutations in all, and $C_m$ acts transitively on $M$, i.e.
for every pair of elements $a,b \in M$, there is an element of $C_m$
that maps $a$ to $b$.
Let $C'_m \subset P_n$ be the permutations in $P_n$ that leave $N\setminus M$
fixed, and restricted to $M$ yield one of the permutations in $C_m$.
There is a natural one-to-one correspondence between $C'_m$ and $C_m$.
We claim that the $m$ permutations $C'_m$ intersect every permutation in
$P_n$.
Suppose, to the contrary, that there exists a permutation $p \in P_n$
that does not intersect any permutation in $C'_m$. Since $C'_m$ acts
transitively on $M \subset N$ the permutation $p$ cannot send any element of
$M$ to any other element of $M$, therefore it must send all the
elements of $M$ to $N\setminus M$, but since $N\setminus M$ has $n - m$ elements and $m > n - m$, this is impossible
by the pigeonhole principle. Therefore such a permutation cannot
exist, and the permutations in $C'_m$ intersect every permutation in
$P_n$.
For $n = 2010$ we get $m = 1006$, which is the required special
case of the general result above.