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 \neq 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.