Find all natural numbers $n> 1$ for which the following applies:
The sum of the number $n$ and its second largest divisor is $2013$.

(R. Henner, Vienna)
1. Let \( n \) be a natural number greater than 1, and let \( p \) be the smallest prime divisor of \( n \). The second largest divisor of \( n \) is then \( \frac{n}{p} \).
2. According to the problem, the sum of \( n \) and its second largest divisor is 2013. Therefore, we have:
   \[
   n + \frac{n}{p} = 2013
   \]
3. Factor out \( n \) from the left-hand side:
   \[
   n \left(1 + \frac{1}{p}\right) = 2013
   \]
4. Simplify the expression inside the parentheses:
   \[
   n \left(\frac{p+1}{p}\right) = 2013
   \]
5. Solve for \( n \):
   \[
   n = \frac{2013p}{p+1}
   \]
6. Since \( n \) must be a natural number, \( \frac{2013p}{p+1} \) must also be a natural number. This implies that \( p+1 \) must be a divisor of 2013.
7. Factorize 2013:
   \[
   2013 = 3 \times 11 \times 61
   \]
   The divisors of 2013 are: 1, 3, 11, 33, 61, 183, 671, and 2013.
8. We need \( p+1 \) to be one of these divisors. Therefore, we check the possible values for \( p \):
   \[
   p+1 = 3 \implies p = 2
   \]
   \[
   p+1 = 11 \implies p = 10 \quad (\text{not a prime})
   \]
   \[
   p+1 = 61 \implies p = 60 \quad (\text{not a prime})
   \]
   The only valid prime \( p \) is 2.
9. Substitute \( p = 2 \) back into the equation for \( n \):
   \[
   n = \frac{2013 \times 2}{2+1} = \frac{4026}{3} = 1342
   \]
10. Verify the solution:
    \[
    1342 + \frac{1342}{2} = 1342 + 671 = 2013
    \]
    The condition is satisfied.

The final answer is \( \boxed{ n = 1342 } \).