343. A nine-digit number divisible by 37 was split into two parts and these parts were swapped. Is the resulting nine-digit number always divisible by 37?
Let the original number be

$$
m=\overline{a_{1} a_{2} \ldots a_{9}}
$$

and it is divided into parts

$$
x=\overline{a_{1} a_{2} \ldots a_{k}}, \quad y=\overline{a_{k+1} a_{k+2} \ldots a_{9}} \quad(1 \leqslant k \leqslant 9)
$$

Then

$$
m=x \cdot 10^{9-k}+y
$$

By the condition

$$
m=37 b, \quad x \cdot 10^{9-k}+y=37 b \quad(b \in N)
$$

from which

$$
y=37 b-x \cdot 10^{9-k}
$$

Write the new nine-digit number:

$$
n=\overline{a_{k+1} a_{k+2} \ldots a_{9} a_{1} a_{2} \ldots a_{k}}=y \cdot 10^{k}+x
$$

Form the difference $n-m$ to determine whether it is always divisible by 37:

$$
\begin{gathered}
n-m=\left(y \cdot 10^{k}+x\right)-\left(x \cdot 10^{9-k}+\mathrm{y}\right)= \\
=\left(37 b-x \cdot 10^{9-k}\right) \cdot 10^{k}+x-x \cdot 10^{9-k}-37 b+x \cdot 10^{9-k}= \\
=37 b \cdot 10^{k}-37 b-x \cdot\left(10^{9}-1\right)=\left(37 b \cdot 10^{k}-37 b\right)-x \cdot 999999999 .
\end{gathered}
$$

The number 999999999 is divisible by 111, and therefore by 37.

It turns out that the difference $n-m$ is divisible by 37. In addition, the subtrahend $m$ is divisible by 37. Therefore, the minuend $n$ is also divisible by 37.

Answer: always.