, 2023

First we prove that she can always obtain 167 points. Without loss of generality, assume there are at least $666/2 = 333$ wheat fields.

Consider the following operation: let the hen start from the field at the front of the row. If she is on a wheat field, she immediately lays an egg and then advances one field; otherwise, she first advances one field, lays an egg, and then advances one more field. Note that this strategy guarantees that every wheat field has an egg, so $W \ge 333$. Moreover, for every rice field with an egg, it is guaranteed that there is a rice field without an egg preceding it, hence $R \le [333/2] = 166$, so $W - R \ge 333 - 166 = 167$.

Now we prove that $S = 167$ is the maximum. Consider the following arrangement (where R denotes a rice field, W denotes a wheat field):
$$\{R\}, \{W, W\}, \{R, R\}, \{W, W\}, \dots, \{R, R\}, \{W\}$$
Note that if the hen lays eggs in $k$ of the W-brackets above, then she can lay eggs in at most $2k$ wheat fields, and must lay eggs in at least $k-1$ rice fields, so $W-R \le 2k-(k-1)=k+1$. This means that when $k < 167$, we have $W-R \le 167$.

It remains to consider the case $k = 167$. Note that in the above arrangement, there are 167 W-brackets and 167 R-brackets each. This means that when $k = 167$, the W-brackets with eggs must include the rightmost $\{W\}$, so at most $2k-1$ wheat fields have eggs, and thus $W-R \le (2k-1)-(k-1) = k = 167$. This completes the proof.