I fear the problem setters blew it big with this one. They forgot to ask for a strategy that works for any $n\times n$ square board! Since the common meaning for chessboard is an $8\times 8$ board, let us just notice that Serezha is forced to win, whatever he does, on any $n\times n$ board with $n$ even. This is obviously so because, any times he puts a chip, it is on a column where Tanya just put her chip; thus after Serezha plays, each column will still contain an even number of empty squares, so that if Tanya can play, he can also play at next turn. (I know this was the interpretation of at least some competitors, since someone complained about this problem, where "any strategy works, so what the h..k of a strategy is that?".) Of course, the setters meant it for any $n>1$. Then, for odd $n$, Serezha indeed needs a strategy, otherwise he may lose. One strategy (the strategy?) is, after he randomly makes his first move, to always play on the other row that contains chips than the one Tanya just played on. This ensures all chips always occupy only two rows, so whenever Tanya can play, so can Serezha, at his next turn. What a pity

For any $n\times n$ chessboard, the second player (who always has to stay in the same column) can win. The winning strategy is to always play in the same two rows, that is, in the row of the starting move and the row of the first counter-move. Then the first player can never leave these rows, and in every move must touch a new column. The second player always reacts in the just opened column, and then this column is never used again.

Let us shoot this fly with a big gun. I will make a quite detailed presentation, in order to convince myself it's working ... Let me first start with some trivial observations. $\bullet$ The game cannot end up in a tie. Trivially so, because eventually the empty squares run out; the game cannot last for more than $n^2$ moves. A simple consequence is that either Tanya or Serezha must have a winning strategy. (The weak form of Zermelo's theorem, stating that in any finite two-person game of perfect information in which the players move alternatively and in which chance does not affect the decision making process, if the game cannot end in a draw, then one of the two players must have a winning strategy.) $\bullet$ Say the board is a square $ABCD$ with $AB$ horizontal, such that $A$ is its NW corner and $B$ is its SW corner. Say Tanya sits having side $BC$ in front of her, while Serezha sits having side $AB$ in front of him. Now, from her second turn, Tanya must move on the row Serezha made his last play, while, from his first turn, Serezha must move on the column Tanya made her last play; from his point of view, he moves along a row of the board, as he sees it. Under this interpretation, the game has a definite symmetry in what its rules are concerned. $\bullet$ The first two moves of Tanya and the first one of Serezha are irrelevant; they create a similar situation on the board. Say then that Tanya starts by putting a chip on the corner $A$ square, denoted by $T_1$ in the sequel. Assume now that Tanya has a winning strategy. According to the above, any strategy of hers starts with two random legal moves; we just fixed the first one to be $T_1$. Now, in the course of her (winning) strategy, the square $T_1$ is out of play, having been occupied. But then Serezha can steal Tanya's strategy. (John Nash's non-constructive strategy-stealing argument.) Serezha starts by making his first move on the corner $B$ square, denoted by $S_1$ in the sequel, in effect starting with the second move (shown to be arbitrary) of Tanya's (winning) strategy. What Tanya sees now is precisely what Serezha saw when he made his move, only that for her one more other square of the board than $S_1$ is out of any further play, namely $T_1$. This means that indeed Serezha can successfully steal Tanya's strategy; he will not be hindered by not being able to make a winning move dictated by this strategy, because that square is out of play, since in fact he plays with one less unavailable square then those Tanya has. Thus Serezha must win, clearly a contradiction. So Tanya has no winning strategy. So Serezha must have a winning strategy.

Well for $n=1$ Tanya wins. My solution is to just look at the top two rows. WLOG that $T$ starts in the top left. Now, $S$ can tile the top two rows with $n$ vertical dominos, and when $T$ plays on one square of the domino you can just have $S$ play on the other square. In this way after both players go $n$ times, $T$ is out of moves and $S$ wins when $n \ge 2$. My interpretation of chessboard was $n$ by $n$ for $n \ge 1$. I have noticed Russian olympiads tend to be a little careless with their quantifiers, and they make the statements very concise but sometimes also slightly confusing.

The second player wins.He just makes the game play between the row it started and an adjacent row.It is Tanya that uses a new column and hence loses after some time.

Serezha always has a winning strategy. Label the rows and columns $1$ through $8$ in that order, and set up our coordinate axes as usual. If Tanya's first chip is placed in $(x_1, y_1)$, then we let Serezha place their first chip in $(x_1, y_2)$ where $y_1 \ne y_2$. Now, it's clear that Tanya's second chip must be placed in $(x_2, y_2)$ for some $x_2$ distinct from $x_1$. Afterwards, we let Sherezha place their second chip in $(x_2, y_1)$. Because Tanya's next chip must be in row $y_1$, we can repeat the process shown above with different values for $x_1, x_2$. It's easy to see that after four iterations, row $y_1$ will be completely filled. Because Serezha's last chip must've been placed in row $y_1$, however, it's clear that Tanya's next chip must also be placed in row $y_1$. But this leads to a contradiction, so we can conclude Serezha always has a winning strategy. $\blacksquare$ Remark: The key insight is to notice that Serezha can force the game to consist of repeatedly drawing rectangles.