335. Find all pairs of natural numbers such that the sum of each with 1 is divisible by the other number.
Let's denote the required numbers by $a$ and $b$. According to the problem,

$$
(a+1) \div b, \quad (b+1) \div a.
$$

We will prove that the numbers $a$ and $b$ are coprime: $\text{GCD}(a, b) = 1$.

Indeed, if $\text{GCD}(a, b) = d$, then

$$
(a+1) \div b, \quad b \div d \Rightarrow (a+1) \div d
$$

But the first term $a$ of the sum $a+1$ is also divisible by $d$, so the second term is divisible by $d: 1 \div d$, which implies $d = 1$.

In this case, the system of the two divisibilities written above is equivalent to the following:

$$
(a+b+1) \div ab
$$

Indeed, the divisibilities (1) imply the divisibility (2), and the divisibility (2) implies each of the divisibilities (1).

(Note that the equivalence mentioned here is valid only when the natural numbers $a$ and $b$ are coprime.)

Let $a \leqslant b$. This restriction does not hinder the solution, as the divisibilities (1) are symmetric with respect to $a$ and $b$. Since the sum $a+b+1$ is divisible by the product $ab$, we have:

$$
ab \leqslant a+b+1 \leqslant b+b+b = 3b
$$

It turns out that $ab \leqslant 3b$. Therefore, $a \leqslant 3$, which means $a \in \{1, 2, 3\}$. Let's consider three cases.

1) Suppose $a = 1$.

Then the divisibility $(a+1) \div b$ becomes $2 \div b$. Hence, $b = 1$ or $b = 2$. In both cases, the second divisibility $(b+1) \div a$ is satisfied.

2) Suppose $a = 2$.

Then $3 \div b$. Therefore, $b = 1$ or $b = 3$. But the first possibility must be discarded because the inequality $a \leqslant b$ must hold, and the second gives the correct divisibility $(3+1) \div 2$.

3) Suppose $a = 3$.

We get $4 \div b$. Out of the three divisors 1, 2, and 4 of the number 4, we need to keep only the last one: $b = 4$. But then the divisibility $(b+1) \div a$ does not hold.

Answer: 1, 1; 1, 2; 2, 3.