1・192 Konya and Betya divide $2 n+1$ nuts, $n \geqslant 2$, and each wants to get as many as possible. Assume there are three ways to divide (each way has three steps).

First step: Betya divides all the nuts into two parts, each part having no less than two nuts.
Second step: Konya then divides each part into two parts, each part having no less than 1 nut (the first and second steps are the same for all three methods).

Third step: In the first method, Konya takes the largest and the smallest part; in the second method, Konya takes the two middle parts; in the third method, Konya either takes the largest and the smallest parts or the two middle parts, but according to the rules, he must give one nut to Betya.

Determine: which method is most advantageous for Konya, and which method is least advantageous for him?
[Solution] The most favorable method for Konya is the first one, and the least favorable method is the third one.

In fact, after Betia's first move, regardless of the size of the walnut piles (each with $a$ and $b$ walnuts, $a<b$), Konya can always divide the larger pile into two parts, each containing 1 and $b-1$ walnuts, which are also the largest and smallest portions, i.e., in the first method, Konya gets $b \geqslant n+1$ walnuts (after Betia takes $a=n, b=n+1$, he can only get $n+1$ walnuts).

In the second method, after Betia's first move with $a=2 ; b=2 n-1$, the best result is dividing them into $2=1+1, 2 n-1=n-1+n$, and Konya only gets $n$ walnuts.

In the third method, when Betia divides the piles as $a=n, b=n+1$, this method does not allow Konya to get more than $n+1$ walnuts (the sum of the two medium piles and the sum of the largest and smallest piles are always exactly $n$ and $n+1$), but he still has to give up one walnut, so Konya gets at most $n$ walnuts.