Solution:
Let set $B$ be the good subset of $A$ which has the maximum number of elements. We can easily see that the number $1$ does not belong to $B$ since $1$ divides all natural numbers. Based on the property of divisibility, we know that $x$ divides the sum of the remaining numbers if and only if $x$ divides the sum of all numbers in the set $B$. If $B$ has exactly $62$ elements, then $B=\{2,3,4, \ldots, 63\}$, but this set can't be good since the sum of its elements is $2015$ which is divisible by $5$. Therefore $B$ has at most $61$ elements. Now we are looking for the set whose elements do not divide their sum, so the best way to do that is making the sum of elements be a prime number. $2+3+4+\ldots+63=2015$ and if we remove the number $4$, we will obtain the prime number $2011$. Hence the set $B=\{2,3,5,6,7, \ldots, 63\}$ is a good one. We conclude that our number is $61$.