Let $n$ be a natural composite number. For each proper divisor $d$ of $n$ we write the number $d + 1$ on the board.  Determine all natural numbers $n$ for which the numbers written on the board are all the proper divisors of a natural number $m$. (The proper divisors of a natural number $a> 1$ are the positive divisors of $a$ different from $1$ and $a$.)
1. Suppose that $(n, m)$ satisfies the condition. We need to determine all natural composite numbers $n$ for which the numbers $d + 1$ (where $d$ is a proper divisor of $n$) are all the proper divisors of a natural number $m$.

2. Note that the least proper divisor of a composite number $n$ is always a prime number. Let $q$ be the least proper divisor of $n$. Then $q + 1$ is the least proper divisor of $m$.

3. Since $q$ is a prime number and $q + 1$ is also a proper divisor of $m$, $q + 1$ must also be a prime number. The only consecutive prime numbers are $2$ and $3$. Therefore, $q = 2$.

4. Since $2$ is the least proper divisor of $n$, $n$ must be even. Let $n = 2k$ for some integer $k$. The largest proper divisor of $n$ is $n/2 = k$. Therefore, $k + 1$ is the largest proper divisor of $m$.

5. We have $k + 1 = m/3$ because $k + 1$ is the largest proper divisor of $m$. Since $2 \nmid m$, $m$ must be odd. Therefore, $4 \mid n$ (since $n$ must be divisible by $4$ to ensure $k$ is an integer).

6. Let $n = 4k$ for some integer $k$. Then $m = 3(k + 1) = 3(2k + 1)$. 

7. If $k = 1$, then $n = 4 \cdot 1 = 4$ and $m = 3(2 \cdot 1 + 1) = 9$. We need to check if the numbers written on the board for $n = 4$ are proper divisors of $m = 9$:
   - Proper divisors of $4$ are $2$.
   - Numbers written on the board are $2 + 1 = 3$.
   - Proper divisors of $9$ are $3$.
   - Therefore, $(n, m) = (4, 9)$ satisfies the condition.

8. Suppose $k \geq 2$. Since $k$ is a proper divisor of $n$, $k + 1$ must be a proper divisor of $m = 3(2k + 1)$. This implies $k + 1 \mid 3(2k + 1)$.

9. For $k = 2$, $n = 4 \cdot 2 = 8$ and $m = 3(2 \cdot 2 + 1) = 15$. We need to check if the numbers written on the board for $n = 8$ are proper divisors of $m = 15$:
   - Proper divisors of $8$ are $2, 4$.
   - Numbers written on the board are $2 + 1 = 3$ and $4 + 1 = 5$.
   - Proper divisors of $15$ are $3, 5$.
   - Therefore, $(n, m) = (8, 15)$ satisfies the condition.

10. For $k \geq 3$, we need to check if $k + 1 \mid 3(2k + 1)$. This condition is not generally satisfied for $k \geq 3$.

Therefore, the only solutions are $n = 4$ and $n = 8$.

The final answer is $\boxed{4, 8}$.