EDIT: solved.
I haven't solved the problem, but I think I know how to begin.
For reference, "k is not divisible by the square of a prime" is the same as "k is squarefree." Also, the term squareful is the same as nonsquarefree.
The discussion on this post is very relevant. We can compare prime factorisations of the numerator and denominator of $\displaystyle \binom{n}{\lfloor n/2 \rfloor}=\frac{n!}{\lfloor n/2 \rfloor! \cdot \lceil n/2 \rceil!}$. For a particular prime, we are interested in whether the exponent in the numerator minus the exponent in the denominator is greater than or equal to 2, in which case the result is squareful. If there is no prime with this property, then the result is squarefree.
For ease of notation, let $\displaystyle k_1=\lfloor n/2 \rfloor$ and $\displaystyle k_2=\lceil n/2 \rceil$.
We can consider each prime power $\displaystyle p^\alpha$ separately, in which case we are comparing the quantities $\displaystyle \bigg\lfloor \frac{n}{p^\alpha} \bigg\rfloor$ and $\displaystyle \bigg\lfloor \frac{k_1}{p^\alpha} \bigg\rfloor + \bigg\lfloor \frac{k_2}{p^\alpha} \bigg\rfloor$.
I've only done some preliminary hand calculations and pattern searching and have not proven anything so far. With more time, I'm pretty confident a pattern would emerge which could then be proven. If you're not required to write a proof and are allowed to write a computer program, a naive algorithm will run very quickly for an upper bound of 200.
C(71, 35) = 221256270138418389602
C(71,35) - Wolfram|Alpha
Confused...