352. Find a natural number that is equal to the sum of all its preceding natural numbers. Does more than one such number exist, or is there only one?
Let $n$ be the desired natural number; the sum of the numbers preceding this natural number is expressed by the formula

$$
S_{n-1}=\frac{n(n-1)}{2}
$$

According to the problem, $\frac{n(n-1)}{2}=n$, from which $n^{2}-3 n=0$ or $n_{1}=3, n_{2}=0$. The desired number is 3. It is the only one, as the second root of the obtained equation (zero) cannot be considered, since no natural number precedes it.