Problem: Two permutations $a_1,a_2,\dots,a_{2010}$
and $b_1,b_2,\dots,b_{2010}$ of the numbers $1,2,\dots,2010$
are said to intersect if $a_k=b_k$ for some value of $k$ in the range $1\le k\le 2010$.
Show that there exist $1006$ permutations of the numbers $1,2,\dots,2010$
such that any other such permutation is guaranteed to
intersect at least one of these $1006$ permutations.

Solution: A valid choice is the following $1006$ permutations:
\[
\begin{array}{cccc ccc | ccccc}
1 & 2 & 3 & \dotsb & 1004 & 1005 & 1006 & 1007 & 1008 & \dotsb & 2009 & 2010 \\
2 & 3 & 4 & \dotsb & 1005 & 1006 & 1 & 1007 & 1008 & \dotsb & 2009 & 2010 \\
3 & 4 & 5 & \dotsb & 1006 & 1 & 2 & 1007 & 1008 & \dotsb & 2009 & 2010 \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots &
\vdots& \vdots & \vdots & \vdots & \vdots \\
1004 & 1005 & 1006 & \dotsb & 1001 & 1002 & 1003 & 1007 & 1008 & \dotsb & 2009 & 2010 \\
1005 & 1006 & 1 & \dotsb & 1002 & 1003 & 1004 & 1007 & 1008 & \dotsb & 2009 & 2010 \\
1006 & 1 & 2 & \dotsb & 1003 & 1004 & 1005 & 1007 & 1008 & \dotsb & 2009 & 2010 \\
\end{array}
\]
This works. Indeed, any permutation should have one of $\{1, 2, \dots, 1006\}$
somewhere in the first $1006$ positions,
so one will get an intersection.

\begin{remark*}
  In fact, the last $1004$ entries do not matter with this
  construction, and we chose to leave them as $1007, 1008, \dots, 2010$
  only for concreteness.
\end{remark*}
\begin{remark*}
  Using Hall's marriage lemma one may prove that
  the result becomes false with $1006$ replaced by $1005$.
\end{remark*}