Let's consider, which values $n$ can take.

If $n = 2k+1 > 1$, let's consider the following distribution of weights. Pieces $n-1$, $n$ and $n+1$ have weight 1, and other ones – 0. Then after Olesya's first step Andriy can not take non-zero pieces. Until Andriy takes by his turn piece $n-2$ or $n+2$, Olesya just makes her step so, that sector of taken pieces is symmetrical to sector with number 0. If Andriy takes by his turn one of pieces $n-2$ or $n+2$, then it does not matter which other piece he takes, Olesya can take two pieces weighting 1 and win.

If Andriy by his first step takes pieces of zero weight 1 and $2n-1$, Olesya chooses two big pieces with numbers 2 and 3. After this Andriy by his step can take at maximum two more big pieces, and Olesya by her turn – two more big pieces and win.

So, we assume, that Andriy by his first step chooses pieces 1 and 2 and we have the total weight $\frac{99}{100}$. After that Olesya takes pieces 3 and 4 and we have the total weight $\frac{201}{100}$. Andriy by any his step can not take piece $2n-1$, because Olesya takes two more big pieces and wins. So Andriy has
---
to go in direction 4; 5; 6; ..., Olesya takes two neighboring in this direction. No body is beneficial to take a piece with number $2n-1$, because then opponent can take two big pieces at once. So, they take zero pieces until they come to big pieces number $2n-4$, $2n-3$ and $2n-2$. According to the order of steps, Andriy takes big pieces with numbers $2n-3$ and $2n-2$. Olesya by her last turn takes piece number $2n-1$, and by last but one – two pieces with numbers $2n-5$ and $2n-4$, the last of which is big. In total we have, that Olesya's pieces' weight will be: $\frac{201}{100} + \frac{101}{100} = \frac{302}{100}$, and Andriy's: $\frac{99}{100} + \frac{199}{100} = \frac{298}{100}$. So, Olesya wins.

If $n=2$, so by his first step Andriy will choose two the biggest pieces and win.
If $n=4$, we have in total $2n=8$ pieces. Let's show, that Andriy always wins. Here number, which denotes the number of piece, also means its weight in comparisons. Let $(k-1, k, k+1)$ denote the smallest in weight piece, that has corresponding number, $[k-1, k, k+1]$ – the biggest, and $\{k-1, k, k+1\}$ – middle.
We can assume that $1+2 \ge 6+7$. Let Andriy take by his first step $1+2$ and does not win. This means, that Olesya could take set $6+7, 3+7$ or $3+4$ and win. Let's consider all these cases. Case $6+7$ has contradiction. Really, as $1+2 \ge 6+7$, and Andriy chooses $[3, 4, 5]+\{3, 4, 5\}$, and Olesya has $(3, 4, 5)$, so Olesya loses.
If Olesya gets win after step $3+7$, this means, that inequality is true:
$$3+7+(4,5,6) > \frac{1}{2}M > 1+2+[4,5,6]+\{4,5,6\}. \quad (1)$$
Then Andriy by his first step takes not $1+2$, but $6+7$. Then he can take by his next step pair of pieces, including 3 (Olesya will not reach it). Then Andriy will have at least set: $6+7+3+(2,4)$. As
$$6+7+3+(2,4) > 3+7+(4,5,6) > \frac{1}{2}M,$$
Andriy wins.
If Olesya gets win after step $3+4$, this means, that inequality is true:
$$3+4+(5,6,7) > \frac{1}{2}M > 1+2+[5,6,7]+\{5,6,7\}. \quad (2)$$
If Andriy by his first step plays $6+7$, Olesya can not take $1+2$ and $1+5$, because Andriy will take winning combination $3+4+6+7 > 3+4+(5,6,7) > \frac{1}{2}M$. Thus, Olesya can win with combination $4+5$, thus inequality must be true:
$$4+5+(1,2,3) > \frac{1}{2}M > 6+7+[1,2,3]+\{1,2,3\}. \quad (3)$$
If Andriy by his first step will play $1+7$, so Olesya can not play $5+6$ and $2+6$, because then Andriy takes winning combination $3+4+1+7 > 3+4+(5,6,7) > \frac{1}{2}M$. Thus, Olesya must win with combination $2+3$, and thus inequality must be true:
$$2+3+(4,5,6) > \frac{1}{2}M > 1+7+[4,5,6]+\{4,5,6\}. \quad (4)$$
If to add two last ratios, we have contradiction:
$$2+3+4+5+(1,2,3)+(4,5,6) > 1+6+7+7+[1,2,3]+\{1,2,3\}+[4,5,6]+\{4,5,6\}$$
as $2+3 \le [1, 2, 3] + \{1, 2, 3\}$, $4+5 \le [4, 5, 6] + \{4, 5, 6\}$, $(1, 2, 3) \le 1$, $(4, 5, 6) \le 6$, $0 < 7+7$. Thus, Andriy has winning strategy, knowing ratio of weights of pieces.