Let $(a_n)_n\geq 0$ and $a_{m+n}+a_{m-n}=\frac{1}{2}(a_{2m}+a_{2n})$ for every $m\geq n\geq0.$ If $a_1=1,$ then find the value of $a_{2007}.$
1. **Initial Condition and Simplification:**
   Given the functional equation:
   \[
   a_{m+n} + a_{m-n} = \frac{1}{2}(a_{2m} + a_{2n})
   \]
   for every \( m \geq n \geq 0 \), and \( a_1 = 1 \).

2. **Setting \( m = n \):**
   \[
   a_{2m} + a_0 = \frac{1}{2}(a_{2m} + a_{2n}) \implies a_{2m} + a_0 = \frac{1}{2}(a_{2m} + a_{2m}) \implies a_{2m} + a_0 = a_{2m}
   \]
   This simplifies to:
   \[
   a_0 = 0
   \]

3. **Setting \( n = 0 \):**
   \[
   a_m + a_m = \frac{1}{2}(a_{2m} + a_0) \implies 2a_m = \frac{1}{2}a_{2m} \implies a_{2m} = 4a_m
   \]

4. **Inductive Hypothesis:**
   We hypothesize that \( a_n = n^2 \) for all \( n \geq 0 \). We will prove this by induction.

5. **Base Case:**
   \[
   a_0 = 0^2 = 0 \quad \text{and} \quad a_1 = 1^2 = 1
   \]
   These are given and match our hypothesis.

6. **Inductive Step:**
   Assume \( a_k = k^2 \) for all \( k \leq n \). We need to show \( a_{n+1} = (n+1)^2 \).

7. **Using the Functional Equation:**
   For \( m = n+1 \) and \( n = 1 \):
   \[
   a_{(n+1)+1} + a_{(n+1)-1} = \frac{1}{2}(a_{2(n+1)} + a_2)
   \]
   Simplifying:
   \[
   a_{n+2} + a_n = \frac{1}{2}(a_{2n+2} + a_2)
   \]
   Using \( a_{2k} = 4a_k \):
   \[
   a_{2(n+1)} = 4a_{n+1} \quad \text{and} \quad a_2 = 4a_1 = 4
   \]
   Substituting these into the equation:
   \[
   a_{n+2} + a_n = \frac{1}{2}(4a_{n+1} + 4)
   \]
   \[
   a_{n+2} + a_n = 2a_{n+1} + 2
   \]
   Using the inductive hypothesis \( a_n = n^2 \) and \( a_{n+1} = (n+1)^2 \):
   \[
   a_{n+2} + n^2 = 2(n+1)^2 + 2
   \]
   \[
   a_{n+2} + n^2 = 2(n^2 + 2n + 1) + 2
   \]
   \[
   a_{n+2} + n^2 = 2n^2 + 4n + 2 + 2
   \]
   \[
   a_{n+2} + n^2 = 2n^2 + 4n + 4
   \]
   \[
   a_{n+2} = 2n^2 + 4n + 4 - n^2
   \]
   \[
   a_{n+2} = n^2 + 4n + 4
   \]
   \[
   a_{n+2} = (n+2)^2
   \]
   Thus, by induction, \( a_n = n^2 \) for all \( n \geq 0 \).

8. **Finding \( a_{2007} \):**
   \[
   a_{2007} = 2007^2
   \]

The final answer is \( \boxed{2007^2} \)