Hmm a bit on the easy side I guess, but still nice.
1. (Russia 2011, 11.4) Ten cars, which do not necessarily start at the same place, are all going one way on a highway which does not loop around. The highway goes through several towns. Every car goes with some constant speed in those towns and with some other constant speed out of those towns. For different cards these speeds can be different. 2011 flags are put in different places next to the highway. Every car went by every flag, and no car passed another right next to any of the flags. Prove that there are at least two flags at which all cars went by in the same order.
Let

denote the time it takes for car

to get to the first flag (choose a starting time arbitrarily). Suppose car

takes time per unit distance

when in and out of town, respectively; let

(not both zero) be the distances from flag

to flag

along in and out of town routes. Then car

takes

to get to flag
. If

for some
, then we must have

for all

(or else we violate the "no passing at flags" rule), whence car

is essentially a "clone" of car
. Since clones are ranked the same at each flag, we can remove all clones to get (WLOG)

pairwise distinct cars (WLOG cars

through
). Then

for all

and
.
Now note that for
, 
iff
![\[(x_k,y_k)\in S_{i,j}=\{(x,y)\mid L_{i,j}(x,y)=(a_i-a_j)x+(b_i-b_j)y+(t_i-t_j)>0\},\]](//latex.artofproblemsolving.com/f/9/f/f9f56603a6e3049e2438f5de19aa67b1403e9a91.png)
and

iff
. Graphing

in the Cartesian plane for

(note the symmetry between

and
), we get at most

(where

is the number of lines) distinct regions (well-known, e.g.
this). But

and

are in the same region iff the orderings at flags

are equal, so we're done by pigeonhole.
Comment: This is not ridiculous or anything, but doing the 1-d case helps a bit. Indeed, this generalizes to any number of dimensions, and the bound should be basically tight and not hard to find.
2. (Russia 2002, 11.8) Show that the numerator of
Outline: Bound denominators using powers of

and primes from

to
. Consider

for
, and note that
, so consider

as well. Subtracting dudes, get a contradiction.
Alternatively, use the stronger result of d'Aurizio mentioned
here.
Edit: Actually, this paper looks pretty incorrect...