TITLE: likely open number theory problem: finite sum of $\zeta(2)$ equal to a square of rationals
QUESTION [12 upvotes]: Which $n$ can let $S=1+\frac14+\frac19+\cdots+\frac1{n^2}$ be a square of a rational number? Obviously, $1$ and $3$ work, but how to prove they are the only ones?
I think this problem is really hard. I have asked many professional number theorists in a top-5 math department in US, and none of them can give any clue. I am thinking this could be an open question.
If you cannot solve this, try this: let $S=1+\frac12+\frac13+\cdots+\frac1{n}$. Which n can make this an integer. Prove your result. This is proposed by a famous mathematician Shing Tung Yau for a college math competition. 
Hopefully someone can give a clue to No.1. I suggested if you cannot solve the first one but would like to see an answer, rate this problem up so that it may have a better chance to bring in more experts for a solution. Thanks!
It is also appreciated if you would like to post your thoughts on Yau's problem!

REPLY [2 votes]: I've verified there are only 2 cases for the first 100000 partial sums.
Here's the Mathematica code I used:
Count[
  ParallelTable[Sqrt[HarmonicNumber[r, 2]] ∈ Rationals, {r, 100000}],
  True
]

(* 2 *)