TITLE: $x$ 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.
QUESTION [6 upvotes]: 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.

REPLY [3 votes]: One can say slightly more, not only two of those $a_i$ are equal, actually two of them equal $x$.
Clearly if two consecutive running averages are equal to $x$, then the term added in the second running average was equal to $x$. One obviously cannot select $51$ out of $100$ running averages without this happening at least once (every selected average except the last would block its successor from selection, which gives $50\times2+1$ selected or blocked averages, one too many). If it actually happens more than once, then we have got our two instances of $a_i=x$. But if one wants it to happen just once, then  by a similar argument, every one of the $100$ running averages is either selected or the successor of a selected average, the two conditions being satisfied simultaneously just once. But that means the very first running average was selected, making $a_1=x$, which together with the instance $a_i=x$ for the index $i$ that was both selected and the successor of a selected index gives our two instances.
One can avoid the case distinction by adding a running average $0$ declared to be equal to $x$ (with which the argument "two consecutive averages equal to $x$ implies the second $a_i=x$" remains valid), and argue for at least two pairs of consecutive chosen averages. This is essentially the answer by Eric Wong. I just wanted to show that one arrives at the conclusion also using only rather fantasy-less reasoning.