Solution:
First we divide both sides of the equation by $2$ and get: $100 m + 3 n = 1003$.

Since $m$ and $n$ are natural numbers we immediately get that $m \leq 10$.

Since $3 n$ is divisible by $3$ and $1003$ gives remainder $1$ upon division by $3$, we conclude that $100 m$ must also give the remainder $1$ upon division by $3$.

Since $100 m = 99 m + m$ and $99 m$ is divisible by $3$ we see that $m$ must give remainder $1$ when divided by $3$.

Thus $m$ has to be one of the numbers $1, 4, 7, 10$.

Corresponding $n$'s are $301, 201, 101$ and $1$, respectively.