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 $\boxed{ k = 2 }$.