Here is an answer via maximum likelihood: 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.