Define $m(n)$ to be the greatest proper natural divisor of $n\in \mathbb{N}$. Find all $n \in \mathbb{N} $ such that $n+m(n) $ is a power of $10$.

[i]N. Agakhanov[/i]
1. Define \( m(n) \) to be the greatest proper natural divisor of \( n \in \mathbb{N} \). We need to find all \( n \in \mathbb{N} \) such that \( n + m(n) \) is a power of 10, i.e., \( n + m(n) = 10^k \) for some \( k \in \mathbb{N} \).

2. Note that \( n \geq 2 \) because the smallest natural number \( n \) is 1, and its greatest proper divisor is 0, which does not satisfy the condition \( n + m(n) = 10^k \).

3. Consider the case where \( n \) is even. If \( n \) is even, then \( m(n) = \frac{n}{2} \). Substituting this into the equation, we get:
   \[
   n + \frac{n}{2} = 10^k \implies \frac{3n}{2} = 10^k \implies 3n = 2 \cdot 10^k
   \]
   Since \( 2 \cdot 10^k \) is even, \( 3n \) must be even, which implies \( n \) must be even. However, \( 3n = 2 \cdot 10^k \) implies \( n = \frac{2 \cdot 10^k}{3} \), which is not an integer unless \( 10^k \) is divisible by 3. This is a contradiction because \( 10^k \) is never divisible by 3. Therefore, \( n \) cannot be even.

4. Now consider the case where \( n \) is odd. Let \( n = l \cdot m(n) \) where \( l \in \mathbb{N} \) and \( l \geq 2 \). Then we have:
   \[
   n + m(n) = l \cdot m(n) + m(n) = (l + 1) \cdot m(n) = 10^k
   \]
   Since \( n \) is odd, \( m(n) \) must also be odd. Therefore, we can write \( m(n) = 5^x \) and \( l + 1 = 2^a \cdot 5^b \) where \( x, a, b \in \mathbb{N} \) and \( x + a + b = k \).

5. If \( l > 5 \), then \( m(n) \) would not be the greatest proper divisor of \( n \). Therefore, \( l \leq 5 \). This implies \( 2^a \cdot 5^b \leq 6 \). The only possible values for \( a \) and \( b \) that satisfy this inequality are \( a = 1 \) and \( b = 0 \), which gives \( l + 1 = 2 \), or \( a = 0 \) and \( b = 1 \), which gives \( l + 1 = 5 \).

6. For \( l + 1 = 2 \), we have \( l = 1 \), which is a contradiction because \( l \geq 2 \).

7. For \( l + 1 = 5 \), we have \( l = 4 \). Substituting \( l = 4 \) into the equation, we get:
   \[
   (4 + 1) \cdot m(n) = 10^k \implies 5 \cdot m(n) = 10^k
   \]
   This implies \( m(n) = 2 \cdot 10^{k-1} \). Since \( m(n) \) must be odd, the only possible value for \( k \) is 2. Therefore, \( m(n) = 2 \cdot 10^{2-1} = 20 \), which is not odd. This is a contradiction.

8. Therefore, the only possible value for \( k \) is 2, and we have:
   \[
   (3 + 1) \cdot m(n) = 10^2 \implies 4 \cdot m(n) = 100 \implies m(n) = 25
   \]
   Since \( m(n) = 25 \), we have \( n = 3 \cdot 25 = 75 \).

Thus, the only solution is \( n = 75 \).

The final answer is \( \boxed{75} \).