A pen costs $11$ € and a notebook costs $13$ €. Find the number of ways in which a person can spend exactly $1000$ € to buy pens and notebooks.
To solve the problem, we need to find the number of non-negative integer solutions \((x, y)\) to the equation:
\[ 11x + 13y = 1000 \]

1. **Express \(x\) in terms of \(y\):**
   \[ x = \frac{1000 - 13y}{11} \]
   For \(x\) to be an integer, \(1000 - 13y\) must be divisible by 11.

2. **Determine the values of \(y\) for which \(1000 - 13y\) is divisible by 11:**
   We need to solve:
   \[ 1000 - 13y \equiv 0 \pmod{11} \]
   Simplifying, we get:
   \[ 1000 \equiv 13y \pmod{11} \]
   Since \(1000 \equiv 10 \pmod{11}\) and \(13 \equiv 2 \pmod{11}\), we have:
   \[ 10 \equiv 2y \pmod{11} \]
   Solving for \(y\), we find:
   \[ 2y \equiv 10 \pmod{11} \]
   Multiplying both sides by the modular inverse of 2 modulo 11, which is 6 (since \(2 \cdot 6 \equiv 1 \pmod{11}\)), we get:
   \[ y \equiv 10 \cdot 6 \pmod{11} \]
   \[ y \equiv 60 \pmod{11} \]
   \[ y \equiv 5 \pmod{11} \]

3. **General solution for \(y\):**
   The general solution for \(y\) is:
   \[ y = 5 + 11k \]
   where \(k\) is a non-negative integer.

4. **Determine the range of \(k\) such that \(x\) is non-negative:**
   Substitute \(y = 5 + 11k\) into the expression for \(x\):
   \[ x = \frac{1000 - 13(5 + 11k)}{11} \]
   \[ x = \frac{1000 - 65 - 143k}{11} \]
   \[ x = \frac{935 - 143k}{11} \]
   \[ x = 85 - 13k \]
   For \(x\) to be non-negative:
   \[ 85 - 13k \geq 0 \]
   \[ 85 \geq 13k \]
   \[ k \leq \frac{85}{13} \]
   \[ k \leq 6.538 \]
   Since \(k\) must be an integer, the possible values of \(k\) are \(0, 1, 2, 3, 4, 5, 6\).

5. **Count the number of solutions:**
   There are 7 possible values for \(k\), hence there are 7 solutions.

The final answer is \(\boxed{7}\).