Determine all pairs $(x,y)$ of natural numbers, such that the numbers $\frac{x+1}{y}$ and $\frac{y+1}{x}$ are natural.
To determine all pairs \((x, y)\) of natural numbers such that the numbers \(\frac{x+1}{y}\) and \(\frac{y+1}{x}\) are natural, we start by setting:
\[
\frac{x+1}{y} = k \quad \text{and} \quad \frac{y+1}{x} = m
\]
where \(k\) and \(m\) are natural numbers. This implies:
\[
x + 1 = ky \quad \text{and} \quad y + 1 = mx
\]

We can rewrite these equations as:
\[
x = ky - 1 \quad \text{and} \quad y = mx - 1
\]

Substituting \(x = ky - 1\) into \(y = mx - 1\), we get:
\[
y = m(ky - 1) - 1
\]
\[
y = mky - m - 1
\]
\[
y + m + 1 = mky
\]
\[
y(1 - mk) = - (m + 1)
\]

This equation must hold for natural numbers \(x\) and \(y\). We will consider different cases for \(k\) and \(m\).

### Case 1: \(k > 1\) and \(m > 1\)
If both \(k\) and \(m\) are greater than 1, then:
\[
y(1 - mk) = - (m + 1)
\]
Since \(1 - mk\) is negative (because \(mk > 1\)), \(y\) must be negative, which is not possible since \(y\) is a natural number. Therefore, there are no solutions in this case.

### Case 2: \(k > 1\) and \(m = 1\)
If \(k > 1\) and \(m = 1\), then:
\[
y + 1 = x
\]
Substituting \(x = ky - 1\) into \(y + 1 = x\), we get:
\[
y + 1 = ky - 1
\]
\[
y + 2 = ky
\]
\[
2 = y(k - 1)
\]
Since \(y\) is a natural number, \(y\) must be a divisor of 2. The possible values for \(y\) are 1 and 2.

- If \(y = 1\):
  \[
  2 = 1(k - 1) \implies k - 1 = 2 \implies k = 3
  \]
  Substituting \(k = 3\) into \(x = ky - 1\):
  \[
  x = 3 \cdot 1 - 1 = 2
  \]
  So, \((x, y) = (2, 1)\).

- If \(y = 2\):
  \[
  2 = 2(k - 1) \implies k - 1 = 1 \implies k = 2
  \]
  Substituting \(k = 2\) into \(x = ky - 1\):
  \[
  x = 2 \cdot 2 - 1 = 3
  \]
  So, \((x, y) = (3, 2)\).

### Case 3: \(k = 1\) and \(m > 1\)
This case is symmetric to Case 2. By similar reasoning, we get the pairs \((x, y) = (1, 2)\) and \((x, y) = (2, 3)\).

### Case 4: \(k = 1\) and \(m = 1\)
If \(k = 1\) and \(m = 1\), then:
\[
x + 1 = y \quad \text{and} \quad y + 1 = x
\]
This implies:
\[
x + 1 = y \quad \text{and} \quad y + 1 = x
\]
\[
x + 1 = y \quad \text{and} \quad y + 1 = x
\]
This leads to a contradiction since \(x\) and \(y\) cannot satisfy both equations simultaneously. Therefore, there are no solutions in this case.

### Conclusion
The pairs \((x, y)\) that satisfy the conditions are:
\[
(x, y) = (2, 1), (3, 2), (1, 2), (2, 3)
\]

The final answer is \( \boxed{ (2, 1), (3, 2), (1, 2), (2, 3) } \)