Let $k$ be an integer and $k > 1$. Define a sequence $\{a_n\}$ as follows:

$a_0 = 0$,

$a_1 = 1$, and

$a_{n+1} = ka_n + a_{n-1}$ for $n = 1,2,...$.

Determine, with proof, all possible $k$ for which there exist non-negative integers $l,m (l \not= m)$ and positive integers $p,q$ such that $a_l + ka_p = a_m + ka_q$.
To determine all possible values of \( k \) for which there exist non-negative integers \( l, m \) (with \( l \neq m \)) and positive integers \( p, q \) such that \( a_l + ka_p = a_m + ka_q \), we start by analyzing the given sequence and the equation.

1. **Sequence Definition and Initial Conditions:**
   \[
   a_0 = 0, \quad a_1 = 1, \quad a_{n+1} = ka_n + a_{n-1} \quad \text{for} \quad n \geq 1
   \]

2. **Rewriting the Given Equation:**
   \[
   a_l + ka_p = a_m + ka_q \implies a_l - a_m = k(a_q - a_p)
   \]
   Without loss of generality, assume \( l > m \) and \( q > p \). This implies \( a_l > a_m \) and \( a_q > a_p \).

3. **Analyzing the Sequence:**
   The sequence \( \{a_n\} \) is strictly increasing. We also note that:
   \[
   a_{n+1} - a_n = (k-1)a_n + a_{n-1} \geq a_n - a_{n-1}
   \]
   This implies that the sequence \( \{a_{n+1} - a_n\} \) is increasing.

4. **Case Analysis:**
   We consider three cases based on the relationship between \( q \) and \( l \).

   **Case (i): \( q \geq l + 1 \)**
   \[
   a_l \geq a_l - a_m = k(a_q - a_p) \geq k(a_q - a_{q-1}) \geq k(a_{l+1} - a_l)
   \]
   \[
   (k+1)a_l \geq ka_{l+1}
   \]
   For \( l = 0 \), this inequality is obviously false. For \( l \geq 1 \):
   \[
   (k+1)a_l \geq k^2 a_l + ka_{l-1} > (k+1)a_l
   \]
   This is a contradiction.

   **Case (ii): \( q \leq l - 1 \)**
   \[
   a_l - a_m = k(a_q - a_p) < ka_q \leq ka_{l-1}
   \]
   For \( l = 1 \), \( m = 0 \), the inequality becomes \( 1 < 0 \), which is false. For \( l > 1 \):
   \[
   a_l - a_m < a_l - a_{l-2}
   \]
   \[
   a_m > a_{l-2}
   \]
   \[
   m > l - 2
   \]
   Since \( m < l \), we have \( m = l - 1 \). Thus:
   \[
   a_l - a_{l-1} = k(a_q - a_p)
   \]
   The sequence \( \{a_n\} \) modulo \( k \) is \( \{0, 1, 0, 1, \ldots\} \), so \( k \nmid a_l - a_{l-1} \), which is a contradiction.

   **Case (iii): \( q = l \)**
   \[
   a_l - a_m = ka_l - ka_p
   \]
   \[
   (k-1)a_l = ka_p - a_m \leq ka_p \leq ka_{l-1}
   \]
   For \( l = 1 \), this inequality is false. For \( l > 1 \):
   \[
   (k-1)a_l \leq a_l - a_{l-2}
   \]
   \[
   (k-2)a_l \leq -a_{l-2}
   \]
   The left-hand side is nonnegative, and the right-hand side is nonpositive, so both must be zero. Thus, \( k = 2 \).

5. **Verification for \( k = 2 \):**
   For \( k = 2 \), we take \( l = 2, m = 0, p = 1, q = 2 \):
   \[
   a_2 + 2a_1 = a_0 + 2a_2
   \]
   \[
   2 + 2 \cdot 1 = 0 + 2 \cdot 2
   \]
   \[
   4 = 4
   \]
   This satisfies the equation.

The final answer is \( \boxed{ k = 2 } \).