Cool problem! The answer is no, Vova cannot prevent Pasha from winning. For convenience, we will refer to the two players as V and Sasha from now on. We will now show that Sasha has a winning strategy. Before doing so, we'll define some terminology. Call two positive real numbers $\mathbf{equivalent}$ if they differ by a factor of a power of $3$. For instance, $1$ and $9$ are equivalent, $\pi$ and $\frac{\pi}{27}$ are equivalent. Call two pieces of plasticine $\mathbf{equivalent}$ if their weights are equivalent to each other. Sasha's winning strategy will proceed in two stages. In the first stage, Sasha will generate $100$ equivalent pieces of plasticine. In the second stage, Sasha will use this to then create $100$ pieces of equal masses. To do so, he will first simply cut the plasticine into three pieces. After V's next move, there will then be two pieces of masses $a$ and $b$. Henceforth, he will set aside the piece of mass $b$, and focus his attention on creating pieces with masses equivalent to $b$ or $2b.$ Indeed, the following observation shows that the first step can be easily completed. $\mathbf{Observation.}$ Using a piece of weight $a$, Sasha can cut it in such a way that after V merges together two of the three parts, some remaining part will be equivalent to a piece of mass $b$ or a piece of mass $2b$. $\mathbf{Proof.}$ Simply cut it into pieces of weight $a-2x, x, x$, where $x$ is a sufficiently small real number of the form $\frac{b}{3^k}$ for $k \in \mathbb{N}$ big. $\blacksquare$ Using this process, it's easy to see that Sasha can eventually generate at least $199$ pieces with masses equivalent to $b$ or $2b$. By Pigeonhole, at least $100$ out of these $199$ are equivalent, call their masses $p_1, p_2, \cdots, p_{100}.$ Now, having obtained these $100$ pieces, Sasha will proceed onto stage two. $\mathbf{Observation.}$ If Sasha has a piece of mass $m$, he can cut it in such a way to replace it with a piece of mass $\frac{m}{3}$ on the next move. $\mathbf{Proof.}$ Just cut it into three equal pieces. $\blacksquare$ With this observation, it's easy to see that for any choice of $a_1, a_2, \cdots, a_{100} \in \mathbb{Z}_{\ge 0}$, Sasha can ensure in finite time that there exist $100$ distinct pieces of plasticine with masses $\frac{p_1}{3^{a_1}}, \frac{p_2}{3^{a_2}}, \cdots, \frac{p_{100}}{3^{a_{100}}},$ and so picking suitable $a_1, a_2, \cdots, a_{100}$ will allow Sasha to generate $100$ pieces of equal masses, as desired. Sasha wins! $\square$

Let $a,b$ denote the number of pieces of sizes $1,2$ (the original has size $3^k$ for some $k$ to be determined later). On the first move P splits the big piece into $1,1,\text{big}-2$. From now on, note that in every move V can only do one of the following, assuming besides the largest part, the other parts have sizes $\le 2$: 1. Merge 2 pieces of 1 into a 2. P splits 1,1 from big piece again. 2. Merge 1 and 2 into a 3. P splits 3 into three pieces of 1. 3. Merge 2 pieces of 2 into a 4. P splits 4 into 2,1,1. The net change in $a + b$ is $+1$ in any case, so after $23333333$ moves one of $a,b$ will exceed $100$. Now set $k = \lfloor e^{66666666666} \rfloor$ gives the desired result.

Let the weight of original plasticine 500. Pasha's strategy is taking the heaviest plasticine and cut it into 1, 1 and the remaining. Then every pieces are integer-weighted. Pasha can do this until every pieces has weight less than 3. So there must be some 100 pieces with same weight 1 or 2. (I was confused since this seems to easy for ARMO, but according to official solution I didn't misread the question anyway.)

Solution with Aatman Supkar, Aditya Khurmi, Alan, Anant Mudgal, Arindam, Derek Liu, Ethan Liu, Jeffrey Kwan, Eric Shen (CAN), Leo Zipeng Lin, Maximus Lu, Misheel Otgonbayar, Paul Hamrick, R Correaa, Robin, Rohan Goyal, Sean Li: The answer is that Pasha can win. Let $M$ be a large integer and assume WLOG (by scaling) that there are $M$ grams of plasticine. Claim: Suppose that any time there is a piece with at least $3$ grams of weight, Pasha splits it into three pieces of integer weight. Then this strategy is winning for Pasha if $M$ is large enough. Proof. This algorithm must eventually terminate because after each of Pasha and Vova's moves, the number of pieces has overall increased by $1$. But it can only terminate if all pieces have weight $1$ or $2$. Thus as long as $M > 500$ then Pasha has won. $\blacksquare$

Let us have a $1$ kilogram plasticine initially. We begin with a very intuitive claim. Claim. At some point there are only $1$ and $2$ gram weighted pieces on the table . Proof. Regardless of Vova's play in his turn , Pasha can always make more pieces of plasticine in each turn ,and after every pair of Pasha-Vova plays the number of plasticine pieces strictly increases (at least by $1$), so this eventually happens .$\square$ Now back to the main problem , FTSC assume Pasha cannot win even after this claim ! (i.e. there're less than $100$ pieces of $1 ; 2$ gram weighted pieces). But then the total weight $=1000<1.100+2.100=300$ ,contradiction. $\blacksquare$

Different solution. Pasha wins. Suppose the initial piece of plasticine has weight $1$. We claim the following strategy for Pasha is winning: Suppose all pieces are of the form $2^{-n}$ for $n \in \mathbb{N}$, and let $2^{-x}$ be the heaviest piece. Pasha splits this into pieces with weights $2^{-x - 1}, 2^{-x - 1}, 2^{-x - 2}$. Suppose that, in Vova's previous move, he combined the pieces $2^{-a}$ and $2^{-b}$ for $a > b$. Pasha splits the piece of weight $2^{-a} + 2^{-b}$ into pieces with weights $2^{-a}, 2^{-b - 1}, 2^{-b - 1}$. This maintains the invariant that, after each of Pasha's moves, all pieces are of the form $2^{-n}$ for $n \in \mathbb{N}$. Hence, on each of Vova's moves, he combines two pieces of the form $2^{-a}$ and $2^{-b}$ (possibly where $a = b$); thus, at any time, all pieces are of one of the forms $2^{-a} + 2^{-b}$ and $2^{-n}$, meaning that the above algorithm always produces a unique move for Pasha. Claim: Suppose that Pasha has just moved. There exists $m \in \mathbb{N}$ such that the weight of each piece is in the set $\{2^{-m}, 2^{-m - 1}, 2^{-m - 2}\}$. Proof: We prove this by induction. After Pasha's first move, the pieces on the board have weights $2^{-1}, 2^{-2}, 2^{-2}$. Consider a move made by Pasha. Suppose he splits $2^{-x}$, for some minimal $x$, into $2^{-x - 1}, 2^{-x - 1}, 2^{-x - 2}$, each of which is in the set $\{2^{-x}, 2^{-x - 1}, 2^{-x - 2}\}$. By induction, each of the other pieces has weight in the set $\{2^{-x}, 2^{-x - 1}, 2^{-x - 2}\}$; hence, we are done in this case. Otherwise, suppose Pasha splits $2^{-a} + 2^{-b}$ (where $a > b$) into $2^{-a}, 2^{-b - 1}, 2^{-b - 1}$. Prior to this move, suppose the heaviest piece is $2^{-M}$. By induction, $a, b \in \{ M, M + 1, M + 2 \}$. As $a > b$, we have $M \leq b \leq M + 1$, or $M + 1 \leq b + 1 \leq M + 2$, so $a, b + 1 \in \{ M, M + 1, M + 2 \}$, so we are once again done. $\blacksquare$ To finish, note that the number of pieces increases by $1$ with each pair of moves. Hence, after $300$ pairs, there are $301$ pieces. For some $m$, each of the pieces has weight in the set $\{2^{-m}, 2^{-m - 1}, 2^{-m - 2}\}$. Thus, by Pigeonhole, there are $100$ pieces which have the same weight, whence Pasha wins. $\Box$

Rename Pasha and Vova to Alice and Bob. Suppose there is initially $1$ kilogram of plasticine. Call a real number quirky if it's of the form $2^{-n}$ for some $n\ge 1$. We claim Alice has a winning strategy. Alice's Strategy: Alice starts by splitting $1$ into $(2^{-2},2^{-2},2^{-1})$, and repeats the following algorithm. Suppose Bob combines $2^{-x}$ and $2^{-y}$. If $x=y$, then Bob's move created $2^{-x}+2^{-x}=2^{-x+1}$, which is quirky. If the largest block is $2^{-m}$, then Alice replaces \[ 2^{-m} \quad \to \quad 2^{-m-2}, \ 2^{-m-2}, 2^{-m-1}. \] If $x>y$, then Alice replaces \[ 2^{-x}+2^{-y} \quad \to \quad 2^{-x-1}, \ 2^{-x-1}, \ 2^{-y}. \] Claim: At any point in time right after Alice's move, each block is one of $\{ 2^{-k-2}, \ 2^{-k-1}, \ 2^{-k} \}$ for some $k\ge 1$. Proof: Induct on the number of moves, base case $(2^{-2},2^{-2},2^{-1})$ clear. Assume all numbers are at most two powers of $2$ apart. In the move when $x=y$, all weights are still within $\{2^{-m-2},2^{-m-1},2^{-m}\}$. In the move when $x>y$, the sizes only get closer together. $\blacksquare$ Note that the number of blocks increases by $1$ from one of Alice's moves to the next. Therefore, after $1000>300$ iterations, there will be at least $100$ equal blocks.

A different solution (though, not as elegant as post #5) The answer is NO. Consider a sufficiently small $\epsilon > 0$ (why such $\epsilon$ exists will be mentioned at the end). Suppose the original stone had weight $c$. We keep breaking this into $\epsilon, 2 \epsilon, c - 3\epsilon$. Let $f(x)$ denote numbers of stone with weight $f(x)$. Call a move of Vova bad if the weights he combine are BOTH not in $\{\epsilon,2 \epsilon \}$ and good otherwise. Note that a bad move increases $f(\epsilon) + f(2 \epsilon)$ while a good move increases $f(2 \epsilon) + f(3 \epsilon) + f(4 \epsilon)$. This means if there are $\ge 200$ bad moves, we are done ; on the other hand, is there are $\ge 300$ good moves then we are still done. As every move is either bad or good, so after $500$ moves we must be done. This also shows $1500 \epsilon < c$ is sufficient. $\blacksquare$

NO Rename Pasha and Vova to Chen and Zhang, respectively. Define a "round" as two consecutive turns from Chen and Zhang in that order, and an "anti-round" as two consecutive turns from Zhang and Chen in that order. Also, assume WLOG that both players are immortal. Let $N=199^{1104}$ be a very VERY large number. Notice that after every round, the total amount of pieces increases by $1$. For the first $N$ turns, Chen will randomly select a piece and cut it into pieces of arbitrary sizes. Fix an $\epsilon$ that is less than $\frac{1}{69^{420}}$ the size of any existing pieces. In every following turn, Chen will choose a piece of size $3\epsilon$ or $4\epsilon$ if they exist, and otherwise, randomly select a piece with size $>2\epsilon$ then cut away two parts of size $\epsilon$ from it. Let $w$ be the number of pieces with size $\epsilon$ or $2\epsilon$. I claim that $w$ increases by at least $1$ after every anti-round. It increases by at least $2$ after every single one of Chen's turns, and the only way Zhang can decrease it by $2$ is by merging two pieces of size $\epsilon, 2\epsilon$ or $2\epsilon, 2\epsilon$, to which Chen can respond immediately by splitting the newly created piece with size $3\epsilon$ or $4\epsilon$ to increase $w$ by $3$. By pigeonhole, the problem is hereby solved. $\blacksquare$

solved with NTguy and 1 other