Multilingual hedge fund - Puzzle

Without knowing the hint, I'm not entirely convinced that finding some minimal $x$ such that $\binom{x}{y} \geq 70$ for some $y$, will actually give you the smallest number of languages $x$. But after a bit of experimentation, this seems to be the right way to go. What you are missing is that for a given $x$, you want the $y$-value which gives you the maximum.

It is fairly well-known (from experience with the binomial distribution, anyway) that to maximize $\binom{x}{y}$ for $y$, pick $y = \lfloor x/2 \rfloor$. Also, the values $\binom{x}{y}$ will then grow roughly with $\sqrt{x!}$, so to get something as small as 70 it's easiest to just guess and check small numbers.

And indeed, $\binom{8}{4} = 70$.

I'm a little confused by the hint, but we can arrive at an answer by applying a theorem that's almost tailored to this problem! We want the 70 employees to define a Sperner system over a set of $n$ languages, which is known to require

$$\binom{n}{\lfloor n/2\rfloor} \geq 70$$

and therefore $n \geq 8$.

The individuals' language sets form an antichain. If the total number of languages spoken is $n$, Sperner's theorem states that the largest antichain (of subsets of the set of languages, with respect to the inclusion relation) has size $n \choose{\lfloor \frac{n}{2} \rfloor}$. The lower bound is then ${n \choose{\lfloor \frac{n}{2} \rfloor}} \geq 70$, or $n \geq 8$. The proofs of Sperner's theorem, which are beautiful but quite clever, show that for even $n$ the largest antichain is unique and is the middle-sized subsets. This makes rigorous Andrew Poelstra's answer and shows that the only solution with $8$ languages is to have every individual speaking $4$ of the tongues.

In essence the problem is asking you to intuit Sperner's theorem. The proof of the theorem is (probably) no simpler for $n=8$ than for general even $n$, which is to say that a job interview that asked this and expected a proof would be unsuitable for most positions held by math PhD's.