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.
 A: What it means, is you have to find the smallest number of participants, such that whatever they do there will always be two with the same score.
For example, if there were two possible scores, you would need three people so that some score always repeats.
A: 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.
A: i arrived at the same answer using stars and bars/combination 'with repetition.'  couldn't entirely follow the answer given by OP.
it is simple.  we have $3$ slots with $20$ identical objects to be placed in them and need all possible placements.  it is ${r+(n-1)}\choose r$ where $r$ is the number of identical objects and $n$ the number of slots/containers.  it is $n-1$ because we are counting the 'bars' separating the slots/containers.  an example is, if the slots were $correct|wrong|omitted,$ $*****|*****|**********$, for $5$ correct, $5$ wrong and ten omitted.  so $n=3,r=20$ and ${{r+(n-1)}\choose r}={22\choose 20}=231$ possible test scores.
However, the correct and wrong answers will cancel eachother out resulting in double-counting of scores.  this will occur when there are $1$ or more correct answers and $4$ or more wrong.  so one counting of these will be ${17\choose 15}=136$.  $232-136=95$ possible test scores and therefore we need a minimum of $96$ participants to gaurantee that at least $2$ of them will have the same score.
