, 2019

$333033$
Let $n$ be a positive integer satisfying the conditions of the problem. Then, $n-32$, $n-33$, $n-34$ are multiples of $97$, $100$, $103$, respectively. Then, we have
$$3n+1 = 3(n-32) + 97 = 3(n-33) + 100 = 3(n-34) + 103,$$
from which it follows that $3n + 1$ is a multiple of $97$, $100$, $103$. Any pair from $97$, $100$, $103$ are relatively prime, we see that $3n+1$ must be a multiple of $97 \cdot 100 \cdot 103$.
The smallest positive integer $n$ for which $3n + 1$ is a multiple of $97 \cdot 100 \cdot 103$ is
$$\frac{97 \cdot 100 \cdot 103 - 1}{3} = 333033, \text{ and since we have}$$
$$\begin{aligned} \frac{97 \cdot 100 \cdot 103 - 1}{3} &= 97 \cdot \frac{100 \cdot 103 - 1}{3} + 32 = 100 \frac{97 \cdot 100 \cdot 103 - 1}{3} + 33 \\ &= 103 \cdot \frac{97 \cdot 100 - 1}{3} + 34, \end{aligned}$$
we conclude that $333033$ is the desired answer to the problem.