$x$ 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 
$$A=\{\,k\in\mathbb N_0\mid k\le 100,a_1+\ldots+a_k=kx\,\}.$$ 
Then $A$ contains not only the 51 given indices, but trivially also $0\in A$, i.e. $|A|\ge 52$.
Let $B= (\mathbb N\setminus A)\cap (A+1)$. Then $A\cap B=\emptyset$ and $A\cup B\subseteq \{0,1,\ldots,101\}$ imply $$\tag1|A|+|B|\le 102.$$ For each  $k\in (A+1)\cap A$ we have $1\le k\le 100$ and $$ a_k=(a_1+\ldots+a_k)-(a_1+\ldots a_{k-1})=kx-(k-1)x=x.$$
Now 
$$(A+1)\cap A=(A+1)\setminus(\mathbb N\setminus A)=(A+1)\setminus B $$
and $$ |(A+1)\setminus B|\ge |A+1|-|B|=|A|-|B|\ge 2|A|-102\ge 2$$
Shows that there are at least two indices $k$ with $a_k=x$.
