One, (20 points) A toy factory has a total of 273 labor hours and 243 units of raw materials for producing a batch of teddy bears and kittens. Producing a teddy bear requires 9 labor hours and 12 units of raw materials, with a profit of 144 yuan; producing a kitten requires 6 labor hours and 3 units of raw materials, with a profit of 81 yuan. Under the constraints of labor and raw materials, how should the production of teddy bears and kittens be reasonably arranged to maximize the total profit from producing teddy bears and kittens?
Let the number of bears and cats produced be $x$ and $y$, respectively, with the total profit being $z$ yuan, requiring $m$ labor hours and $n$ units of raw materials. Then,
$$
\left\{\begin{array}{l}
m \leqslant 273, n \leqslant 243, \\
9 x+6 y=m, \\
12 x+3 y=n .
\end{array}\right.
$$

From equations (1) and (2), we get $x=\frac{2 n-m}{15}, y=\frac{4 m-3 n}{15}$.
From equations (1) and (2), it is known that $31 m, 3 \mid n$.
Let $m=3 k, n=3 t\left(k, t \in \mathbf{N}_{+}, k \leqslant 91, t \leqslant\right.$
81). Then
$$
\begin{array}{l}
z=144 x+81 y \\
=144 \times \frac{2 n-m}{15}+81 \times \frac{4 m-3 n}{15} \\
=12 m+3 n=36 k+9 t .
\end{array}
$$

Since $36>9,91>81$, the value of $z$ is maximized when $k$ is as large as possible and $t$ follows suit, while ensuring that $x$ and $y$ are positive integers.
When $k=91$, we have
$$
\begin{array}{l}
x=\frac{2 \times 3 t-3 k}{15}=\frac{2 t-k}{5}=\frac{2 t-91}{5} \\
=\frac{2 t-1}{5}-18 .
\end{array}
$$

Thus, $2 t-1$ is a multiple of 5. Since $2 t-1$ is also an odd number, let $2 t-1=5\left(2 t_{1}+1\right)$.
Then $t=5 t_{1}+3 \leqslant 81$.
Solving, we get $t_{1} \leqslant 15 \frac{3}{5}$.
Therefore, the maximum value of $t_{1}$ is 15.
Hence, the maximum value of $t$ is 78.
At this point, $x=13, y=26$.
Thus, the maximum value of $z$ is
$$
36 \times 91+9 \times 78=3978 \text { (yuan) } \text {. }
$$

Therefore, by producing 13 bears and 26 cats, the total profit can reach its maximum.