$x$ is equal to at least $51$ of $a_1,\frac{a_1+a_2}{2},\ldots,\frac{a_1+a_2+\ldots+a_{100}}{100}$. Prove that $2$ of $a_1,\ldots,a_{100}$ are equal. 
If $x$ is equal to at least $51$ number of the array $a_1, \frac{a_1+a_2}{2},\ldots,\frac{a_1+a_2+\ldots+a_{100}}{100}$, prove that $2$ numbers of the array $a_1,a_2\ldots,a_{100}$ are equal.

That's the way the problem is originally stated. It doesn't tell me whether at least or exactly $2$ numbers of the array $a_1,a_2\ldots,a_{100}$ are equal. 
I've thought of a way to define the problem in a different way: 

If $n_1,n_2,\ldots,n_{100}$ is an array of distinct natural numbers that belong to the interval $[1\:;100]$ and $k\in\mathbb N$, $k\le50$ and
  $\begin{cases}n_1x=a_1+a_2\ldots+a_{n_1}\\n_2x=a_1+a_2\ldots+a_{n_2}\\\ldots\\n_{50+k}x=a_1+a_2\ldots+a_{n_{50+k}}\end{cases}$, prove that $2$ numbers of the array $a_1,a_2\ldots,a_{100}$ are equal.

I have no idea at the moment how we could come up with a proof that $2$ numbers of the array $a_1,a_2\ldots,a_{100}$ are equal. Some ideas would be great. Thanks.

Let $I$ be the set of all $0 \le n \le 100$ such that $x_1 + \cdots + x_n = nx$.
Note that $0$ does belong to $I$, and so do the $51$ (or more) other values stated in the hypothesis, so $|I| \ge 52$.
Since $I$ is a subset of $\{0, \ldots, 100\}$ with size at least $52$, it must contain at least two pairs of consecutive elements.  In other words, there are distinct $m,n$ such that $m, m+1\in I$ and $n, n+1 \in I$.  (One way to see this cleanly is to consider the spacings between the first $52$ elements: there are $51$ spaces totaling at most $100$, so at least two of the spacings must be $1$ rather than $\ge 2$.)
Notice that whenever $r,r+1 \in I$ we have that $a_{r+1} = x$ (since you're confused you may want to double-check that this holds for $r=0$, but it really is the same calculation as any $r$).  Thus with $m$ and $n$ chosen as above, we have $a_{m+1} = a_{n+1} = x$.
