About numbers which have very few little divisors

Assuming I understand your question properly, I believe the product of the first $n$ primes plus $1$ would give a harsh number (let's call it $N$). The only primes with $F(p)\leq N$ would be the first $n$ primes, and they each divide $N-1$.

For the second part, let $P_n$ be the product of the first $n$ primes (or we could let it be $F(p_n)$). Given a number $N$, define $n$ such that $P_n < N < P_{n+1}$. We must have that, since $F(p_k)<N$ for all $1\leq k \leq N$, $P_n|N-1$ (see that this is equivalent to it being harsh). So,

$$N=jP_n+1$$

for some $1\leq j < p_{n+1}$. Thus, the harsh numbers look like this:

$$\cdots,P_n+1,2P_n+1,\cdots,(p_{n+1}-1)P_n+1,P_{n+1}+1,2P_{n+1}+1,\cdots$$

It's hard to index them without knowing the first $n$ primes themselves, but knowing them it's relatively straightforward.

O.K. we can start the answer by seeing that the numbers $n=1+\prod \limits_{p_i \leq f(n)} p_i $ fit your condition, now we just need to see for what functions its true.

Easy to see that if $Max(p_i) \geq n$ then $n$ can not hold true to your condition so it means that $f(n) \geq n$ will give no valid $n$.(Actually we can reduce this to $f(n) \geq \frac{n}{2}$ by same reasoning.).

We know that $p_i \approx i \ln i$ and we want that $\prod \limits_{i=2}^{p} i \ln i \approx n $ and find $p$ using $n$ will give the proper function (An approximation to the function).

Now $\prod \limits_{i=2}^{p} i \ln i \approx e^{\sum\limits_{i=2}^{p} i \ln i}\approx e^{\int\limits_{i=2}^{p} i \ln i}= \frac{1}{4} e^{\frac{1}{4} p^2 (2 \log (p)-1)+1}\approx n $ solving for $p$(which is a symbol for the maximum prime) we get that $p=e^{\frac{1}{2} \left(W\left(\frac{4 (\log (4 n)-1)}{e}\right)+1\right)}$ and since $W(x) \approx \ln x + \ln \ln x$ where $W(x)$ is (Lambert W function) we can approximate to $p \approx \frac{2 \sqrt{\log (4 n)-1}}{\sqrt{\log (4 (\log (4 n)-1))-1}} \approx \frac{2\sqrt{\ln n}}{\sqrt{\ln \ln n}}.$

So when $f(n) \leq \frac{2\sqrt{\ln n}}{\sqrt{\ln \ln n}}$ we will have Infinitely many solutions

For $\frac{2\sqrt{\ln n}}{\sqrt{\ln \ln n}} << f(n) << n$ we could only have finitely many solutions as for your examples $f(n)=\sqrt{n},\ln n$