Question: There is a set a set $S$ of numbers. i.e. $(s\_1, s\_2, s\_3, s\_4, s\_5, ..., s\_n)$. The average of the numbers in the set is $N$. How do I prove that at least one of the numbers in the set is greater than or equal to $N$?

I am thinking that a proof by contradiction is probably the easiest way.

I could do a proof by contradiction by supposing that no $s\_i$ in $S$ is greater than or equal to $N$. But then I am stuck on how to proceed to show that it is False and the original statement is true.

Answer: 
Suppose there are $n$ elements $s\_i \in S$ and the largest is $M$.

Then looking at the mean we have $\dfrac{\sum\_i s\_i} {n}=N$, and each $s\_i \le M$, 

so $ nN=\sum\_i s\_i \le \sum\_i M=nM$ meaning $nN \le nM$ and so $$N \le M$$ i.e. the mean is less than or equal to the maximum. 

No need to look for a contradiction.
Enumerate the elements of $S$ as $S=\{s\_1,\ldots,s\_K\}$. Suppose that it was true that $s\_i<N$ for all $i$. Then,
$$
\sum\_{i=1}^K s\_i<\sum\_{i=1}^K N=K\times N=K\frac{\sum\_{i=1}^K s\_i}{K}=\sum\_{i=1}^K s\_i.
$$
The leftmost term and rightmost term are the same so you have a contradiction.