How many participants required? 
A test consisting of 20 problems is given at a math competition. Each correct answer to each problem gains 4 points; each wrong answer takes away 1 point, and each problem left without an answer gets 0 points. What is the lowest possible number of participants in the competition that is needed, so that at least two of them will get an equal number of points?

I don't see how this is an accurate question even. You can have two participants and get equal number of points.
If you have 3 it works as well. What is the point of this problem?
Thanks!
EDIT:
I think small cases should be good:
C - correct, W - Wrong, M - Omitted
$20C \to 80$.
$19C, 1W, 0M \implies 75$ $19C, 0W, 1M \implies 76$.
$18C, 1W, 1M \implies 71, 18C, 2W, 0M \implies 70, 18C, 0W, 2M \implies 72$.
$17C, 1W, 2M \to 67, 17C, 2W, 1M \to 66, 65, 17C, 0W, 3M \to 68$.
The difference, from the highest to highest is $4$. For each $xC$ there are $(21-x)$ terms. So the series is:
So for $0C$ there are $21-0 = 21$ terms.
Thus, the series is:
$1 + 2 + 3 + ... + 21 = \frac{21 \cdot 22}{2} = 231$.
We need at least $232$ participants.

I understand the problem as follows: What is the minimum number of participants such that we can be sure to see two participants with the same score.
Denote by $c$, $w$ the number of correct, resp., wrong answers of some participant, and by $s$ his score. Then $$c\geq 0,\quad w\geq0,\quad c+w\leq20,\quad s=4c-w\ .$$
Given $c\in[0\ ..\ 20]$ we have $w\in[0\ ..\ 20-c]$, so that $$s=4c-w\in [5c-20\ ..\ 4c]=:J_c\ .$$ The set $S$ of possible scores is then given by
$$S=\bigcup_{c=0}^{20}J_c\ .$$
The $J_c$ are intervals of successive integers, beginning with $J_0=[-20\ ..\ 0]$, moving steadily to the right and becoming shorter. The last interval is the singleton $J_{20}=[80\ ..\ 80]$. As long as the left endpoint of $J_c$ is at most one more than the right endpoint of $J_{c-1}$ we obtain an uninterrupted sequence of integers in $S$. This condition is violated as soon as
$$5c-20\geq 4(c-1)+2\ ,$$
or $c\geq18$. In the transition $17\rightsquigarrow18$ the number $69$ is left out; in the transition $18\rightsquigarrow19$ the numbers $73$ and $74$ are left out, an in the transition $19\rightsquigarrow20$ the numbers $77$, $78$, $79$ are left out. It follows that
$$S=[-20\ ..\ 80]\setminus\{69, 73,74, 77, 78, 79\}\ ,$$
so that $|S|=101-6=95$.
In order to guarantee two contestants with the same score we therefore need $96$ participants.
