Answer: The coins that are in a position that is a perfect square will be tails up, while all other coins will be heads up.
Since all coins were heads up at the start, any quarter that was turned over an even number of times will be heads up at the end of the procedure and any quarter that was turned over an odd number of times will be tails up at the end of the procedure.
The general rule was that in step $n$, exactly those coins in a position that is a multiple of $n$ will be turned over. Thus each coin’s status will be changed exactly once for each positive integer that divides its position. For example, the 12th coin will be turned over in steps 1, 2, 3, 4, 6, and 12 (an even number of times) and will therefore end in a heads up state.
So the problem can be reduced to determining which numbers have an even number of factors and which numbers have an odd number of factors. However, factors come in pairs based on the rule that if $d$ is a factor of $n$, then so is $n/d$. For example, 3 is a factor of 12 and so is $12/3=4$. However, if we ever have $d=n/d$, then this pairing will only correspond to a single factor and the number $n$ will have an odd number of factors. But $d=n/d$, if and only if, $n=d^2$, or equivalently when $n$ is a perfect square. For example, the perfect square has factors: 1 & 16, 2 & 8, and 4. The “unpaired” 4 results in an odd number of factors and thus the 16th coin would end up tails up as would any other coin in a position that is a perfect square.