The answer is yes for $n=2013$ and no for $n=2014$. More generally Ana can complete her task if $n=0 \pmod{3}$ ($n \ge 3$), and Maria can prevent her from doing so if $n \equiv 1 \pmod{3}$ ($n \ge 4$).
Let $n$ be a multiple of 3. Color the coins in 3 colors periodically: 1, 2, 3, 1, 2, 3, ..., 1, 2, 3. The coins of some color $c$ have total value at most $1/3$ of the total sum. We claim that Ana can force Maria to take a coin colored $c$ on every move, which will imply that Ana can ensure at least $2/3$ of the total amount for herself.
Indeed, let $c=1$; then Ana takes the last coin 3 first, then the last coin 2 in the new sequence. Thus Maria must move at a sequence 1, 2, 3, ..., 2, 3, 1, so she has to take a coin 1. Moreover each of her two possible moves yields a sequence either starting or ending with 2, 3. Ana can take these consecutive coins 2, 3 on her next move, thus obtaining a sequence of the kind 1, 2, 3, ..., 2, 3, 1, again. By following the same strategy Ana can ensure that Maria takes away all coin 1, as needed. If $c=2$ Ana takes the first coin 1 and the last coin 3, obtaining 2, 3, 1, ..., 3, 1, 2. Now any move of Maria is 2 and leaves a sequence either starting or ending with 3, 1. Ana removes such two consecutive coins and obtains 2, 3, 1, ..., 3, 1, 2 again. So the pattern repeats: Ana makes sure that takes away all coins 1 and 3.
The case $c=3$ is completely analogous. Here Ana takes the first two coins 1, 2, in succession, yielding a sequence 3, 1, 2, ..., 1, 2, 3. Maria is forced to take a coin 3, after which Ana can restore the pattern 3, 1, 2, ..., 1, 2, 3 by taking two consecutive 1, 2 from an extreme.
Let $n=1 \pmod{3}$. Color the coins periodically
$$1, 2, 3, 1, 2, 3, \dots, 1, 2, 3, 1. \qquad (1)$$
Suppose that all coins 1 are 5 cents and all remaining ones are 1 peso. Then Maria can force Ana to take all coins 1, so that Ana will have less than $2/3$ of the total amount (this is easy to check). If Ana takes two consecutive coins from one extreme of (1), Maria takes the third coin from the same extreme; it is a 2 or a 3. The same pattern (1) occurs. If Ana takes the two extremal 1's in (1), Maria takes the first coin 2 in the resulting sequence. This gives
$$3, 1, 2, 3, \dots, 1, 2, 3. \qquad (2)$$
There are three cases depending on Ana's next move. (a) If Ana takes two coins from the left extreme of (2), Maria takes the coin 2 following them and (2) occurs again. (b) If Ana takes two coins 2, 3 from the right extreme of (2), Maria takes the first coin 3 and (1) occurs again. (c) If Ana takes one coin of each extreme of (2). The resulting sequence ends in a 2. Maria takes this 2 and obtains (1).
Thus Maria ensures that one of the patterns (1) or (2) occurs after each combined move of the two. Eventually the sequence will become 1, 2, 3, 1 or 3, 1, 2, 3. In both cases Maria can make sure that Ana gets the remaining coin(s) 1.