Here is an answer via [maximum likelihood][1]:  choose the population size for which the observed distribution would be most likely. 

Let $n$ be the population size.  There are 77 people encountered, whom we identify as:  1) the person encountered four times, 2) the first person encountered three times, ... 77) the sixtieth person encountered once.   There are therefore ${n \choose 77}$ ways of choosing these people.

The probability of observing this distribution is therefore $$\frac{{n \choose 77}f(1,4,12,60)}{n^{100}}$$ where $f$ Is the number of ways of getting the first person to appear four times, the second to appear three times, ... the seventy-seventh to appear once.  Since all we have to do is maximize the probability, we can ignore $f$. 

Numerically, the probability is maximized for $n=181$.  So 181 is a reasonable estimate for the population size.


  [1]: https://en.m.wikipedia.org/wiki/Maximum_likelihood_estimation