26. Let $n$ bags contain $k_{1}, k_{2}, \cdots, k_{n}$ balls respectively, and two players, A and B, take turns to pick balls. Each time, they can only take from one bag, taking at least 1 and at most $m$ balls. The player who takes the last ball wins. If we denote this problem as $\left\{k_{1}, k_{2}, \cdots, k_{n}\right\}_{m}$, we want to ask: For the problem $\{1,2,3, \cdots, n\}_{3}$, for what value of $n$ does A (the first player) have a winning strategy?
26. First, prove the following lemmas:

Lemma 1 For the problem $\left\{k_{1}\right\}_{3}$, player B has a winning strategy if and only if $4 \mid k_{1}$.
Proof of Lemma 1 If $4 \mid k_{1}$, when $k_{1}=4$, if player A takes $r$ balls $(1 \leqslant r \leqslant 3)$, then player B takes $4-r$ balls to win; when $k_{1}>4$, if player A takes $r$ balls, then player B takes $4-r$ balls, leaving the remaining number of balls as a multiple of 4. This continues, ensuring that player B will win. If $k_{1}=4 r+s(1 \leqslant s \leqslant 3)$, then player A first takes $s$ balls, making the remaining number of balls 0 (when $r=0$, thus player A wins) or still a multiple of 4. From the above analysis, it is clear that player A will eventually win.

Lemma 2 For the problem $\left\{k_{1}, k_{2}\right\}_{3}\left(k_{1} \leqslant k_{2}\right)$, player B has a winning strategy if and only if $4 \mid k_{2}-k_{1}$.
Proof of Lemma 2 When $4 \mid k_{2}-k_{1}$, if each turn of taking balls by player A and player B is considered as one step, then player B just needs to ensure that after each step, the remainder of the number of balls in both pockets when divided by 4 is the same. This way, as long as player A has balls to take, player B will also have balls to take, thus ensuring that player B takes the last ball and wins. If $4 \nmid k_{2}-k_{1}$, let $k_{1}=4 q_{1}+r_{1}, k_{2}=4 q_{2}+r_{2}\left(0 \leqslant r_{1}, r_{2} \leqslant 3\right)$, then there must be $r_{1} \neq r_{2}$. Without loss of generality, assume $r_{1}<r_{2}$, then player A first takes $r_{2}-r_{1}$ balls from the second pocket, making the difference in the number of balls in the two pockets divisible by 4. From the previous analysis, it is clear that player A has a winning strategy.

Lemma 3 If for the problem $\left\{k_{1}, k_{2}, \cdots, k_{n}\right\}_{3}$, player B has a winning strategy, and for the problem $\left\{\lambda_{1}, \lambda_{2}, \cdots \lambda_{m}\right\}_{3}$, player B also has a winning strategy, then for the problem $\left\{k_{1}, \cdots, k_{n}, \lambda_{1}, \cdots, \lambda_{m}\right\}_{3}$, player B has a winning strategy.
Proof of Lemma 3 Player B can divide the problem $\left\{k_{1}, k_{2}, \cdots, k_{n}, \lambda_{1}, \lambda_{2}, \cdots, \lambda_{m}\right\}_{3}$ into two groups, with the number of balls being $k_{1}$, $k_{2}, \cdots, k_{n}$ in the first group, and $\lambda_{1}, \cdots, \lambda_{m}$ in the second group. Player B adopts a winning strategy in each group, combining them into a winning strategy for the problem $\left\{k_{1}, k_{2}, \cdots, k_{n}, \lambda_{1}, \cdots, \lambda_{m}\right\}_{3}$.

Lemma 4 If for the problem $\left\{k_{1}, k_{2}, \cdots, k_{n}\right\}_{3}$, player A has a winning strategy, and for the problem $\left\{\lambda_{1}, \cdots, \lambda_{m}\right\}_{3}$, player B has a winning strategy, then for the problem $\left\{k_{1}, k_{2}, \cdots, k_{n}, \lambda_{1}, \lambda_{2}, \cdots, \lambda_{m}\right\}_{3}$, player A has a winning strategy.
Proof of Lemma 4 Player A can divide the pockets into two groups, with the number of balls being $k_{1}, k_{2}, \cdots, k_{n}$ in the first group, and the rest in the second group. Player A adopts a winning strategy in the first group and a winning strategy for player B in the second group, combining them into a winning strategy for the problem $\left\{k_{1}, k_{2}, \cdots, k_{n}, \lambda_{1}, \cdots, \lambda_{m}\right\}_{3}$.

Returning to the problem $\{1,2, \cdots, n\}_{3}$, when $n=1,2$, by Lemmas 1 and 2, player A has a winning strategy; when $n=3$, player B can always ensure that after the first step (i.e., after player A and player B each take one turn), only two pockets have balls, and the number of balls in these two pockets is the same. Thus, by Lemma 2, player B has a winning strategy at this point; when $n=4$, because for the problems $\{1,2,3\}_{3}$ and $\{4\}_{3}$, player B has a winning strategy, by Lemma 3, for the problem $\{1,2,3,4\}_{3}$, player B has a winning strategy. Also, when $k=1,2,3, \cdots$, for the problem $\{4 k+1,4 k+2,4 k+3,4 k+4\}_{3}$, player B always has a winning strategy. This is because, when player A takes $r$ balls from a pocket with more than 4 balls, player B takes $4-r$ balls from the same pocket, ensuring that the remainder of the number of balls in each pocket when divided by 4 remains unchanged. If player A takes balls from a pocket with no more than 4 balls, player B can adopt the winning strategy for the problem $\{1,2,3,4\}_{3}$. Therefore, the strategy for the problem $\{4 k+1,4 k+2,4 k+3,4 k+4\}_{3}$ is the same as for the problem $\{1,2,3,4\}_{3}$. Since for the problem $\{1,2,3,4\}_{3}$, player B has a winning strategy, for the problem $\{4 k+1,4 k+2,4 k+3,4 k+4\}_{3}$, player B has a winning strategy.

From the above, we can conclude: (1) When $n=4 k(k \geqslant 1)$, player B has a winning strategy, because at this time, the pockets can be grouped into sets of 4 pockets in order of the number of balls, and for each set, player B has a winning strategy. By Lemma 3, for the problem $\left\{1,2,3,4, \cdots, 4 k-3,4 k-2,4 k-1,4 k\right\}_{3}$, player B has a winning strategy; (2) When $n=4 k+1(k \geqslant 0)$, player A has a winning strategy, because when $k=0$, it has been proven that player A has a winning strategy, and when $k \geqslant 1$, it has been proven that for the problem $\{1,2, \cdots, 4 k\}_{3}$, player B has a winning strategy, and by Lemma 1, for the problem $\{4 k+1\}_{3}$, player A has a winning strategy. Therefore, by Lemma 4, for the problem $\{1,2, \cdots, 4 k, 4 k+1\}$, player A has a winning strategy; (3) When $n=4 k+2(k \geqslant 0)$, player A has a winning strategy, because when $k=0$, it has been proven that player A has a winning strategy, and when $k \geqslant 1$, it has been proven that for the problem $\{1,2, \cdots, 4 k\}_{3}$, player B has a winning strategy, and by Lemma 2, for the problem $\{4 k+1,4 k+2\}$, player A has a winning strategy. Therefore, by Lemma 4, for the problem $\{1,2, \cdots, 4 k, 4 k+1,4 k+2\}$, player A has a winning strategy; (4) When $n=4 k+3$, player B has a winning strategy, because when $k=0$, it has been proven that player B has a winning strategy, and when $k \geqslant 1$, if for the problem $\{1,2, \cdots, 4 k+3\}_{3}$, player A has a winning strategy, and for the problem $\{4 k+4\}_{3}$, player B has a winning strategy, then by Lemma 3, for the problem $\{1,2, \cdots, 4 k+3,4 k+4\}_{3}$, player A has a winning strategy, which contradicts the conclusion already proven in (1). Therefore, for the problem $\{1,2, \cdots, 4 k+3\}_{3}$, player B has a winning strategy. In summary, when $n \equiv 1,2(\bmod 4)$, player A has a winning strategy; when $n \equiv 0,3(\bmod 4)$, player B has a winning strategy.