jmc

The positive integers with exactly four positive divisors are the integers of the form $p^3$, where $p$ is a prime, or $p\cdot q$, where $p$ and $q$ are distinct primes. We consider each case:

Suppose that $m=p^3$ for some prime $p$. Then the sum of the divisors of $m$ is $1+p+p^2+p^3.$ For $p=11,$ this value of $m$ is too low, and for $p=13,$ the value of $m$ is too high; therefore, no prime $p$ gives a value of $n$ in the given set.

Therefore, we must have $m=p\cdot q$, for some distinct primes $p$ and $q.$ Then the sum of the divisors of $m$ is $1+p+q+pq$, which we can factor as $(1+p)(1+q)$. First suppose that one of $p$ and $q$ equals $2$; without loss of generality, let $p=2$. Then $(1+p)(1+q)=3(1+q).$ Since $q\ne p=2$, we see that $q$ is odd, and so $1+q$ is even. Thus $3(1+q)$ is divisible by $6,$ so it must be either $2010$ or $2016.$ Trying both cases, we see that both $3(1+q)=2010$ and $3(1+q)=2016$ give a non-prime value of $q.$

If neither $p$ nor $q$ equals $2$, then both are odd primes, so $(1+p)(1+q)$ is the product of two even numbers, which must be divisible by $4.$ The only multiples of $4$ in the given range are $2012$ and $2016$. We have $2012=2^2\cdot 503,$ so the only way to write $2012$ as the product of two even positive integers is $2012=2\cdot 1006.$ But we cannot have $1+p=2$ or $1+q=2$, since $2-1=1$ is not prime. Note that $2016=(1+3)(1+503).$ Since both 3 and 503 are prime, 2016 is nice.

Thus, $\boxed{2016}$ is the only nice number in the given set.