Let $f(x)$ be a polynomial with integer coefficients such that $f(15) f(21) f(35) - 10$ is divisible by $105$. Given $f(-34) = 2014$ and $f(0) \ge 0$, find the smallest possible value of $f(0)$.

[i]Proposed by Michael Kural and Evan Chen[/i]
1. **Understanding the problem**: We need to find the smallest possible value of \( f(0) \) given that \( f(x) \) is a polynomial with integer coefficients, \( f(15) f(21) f(35) - 10 \) is divisible by 105, \( f(-34) = 2014 \), and \( f(0) \ge 0 \).

2. **Divisibility conditions**: Since \( f(x) \) has integer coefficients, we have:
   \[
   f(15) \equiv f(0) \pmod{3} \quad \text{and} \quad f(15) \equiv f(0) \pmod{5}
   \]
   \[
   f(21) \equiv f(0) \pmod{3} \quad \text{and} \quad f(21) \equiv f(0) \pmod{7}
   \]
   \[
   f(35) \equiv f(0) \pmod{5} \quad \text{and} \quad f(35) \equiv f(0) \pmod{7}
   \]

3. **Combining congruences**: From the above, we can deduce:
   \[
   f(15) \equiv f(21) \equiv f(35) \equiv f(0) \pmod{3}
   \]
   \[
   f(15) \equiv f(35) \equiv f(0) \pmod{5}
   \]
   \[
   f(21) \equiv f(35) \equiv f(0) \pmod{7}
   \]

4. **Divisibility by 105**: Given \( f(15) f(21) f(35) - 10 \) is divisible by 105, we have:
   \[
   f(15) f(21) f(35) \equiv 10 \pmod{105}
   \]

5. **Breaking down the modulus**: We need to check the congruences modulo 3, 5, and 7 separately:
   - Modulo 3:
     \[
     f(0)^3 \equiv 10 \pmod{3} \implies f(0)^3 \equiv 1 \pmod{3} \implies f(0) \equiv 1 \pmod{3}
     \]
   - Modulo 5:
     \[
     f(0)^3 \equiv 10 \pmod{5} \implies f(0)^3 \equiv 0 \pmod{5} \implies f(0) \equiv 0 \pmod{5}
     \]
   - Modulo 7:
     \[
     f(0)^3 \equiv 10 \pmod{7} \implies f(0)^3 \equiv 3 \pmod{7}
     \]

6. **Chinese Remainder Theorem**: We need to solve the system of congruences:
   \[
   f(0) \equiv 1 \pmod{3}
   \]
   \[
   f(0) \equiv 0 \pmod{5}
   \]
   \[
   f(0) \equiv 3 \pmod{7}
   \]

7. **Solving the system**:
   - Let \( f(0) = 5k \) for some integer \( k \). Then:
     \[
     5k \equiv 1 \pmod{3} \implies 2k \equiv 1 \pmod{3} \implies k \equiv 2 \pmod{3} \implies k = 3m + 2
     \]
     Thus, \( f(0) = 5(3m + 2) = 15m + 10 \).

   - Now, \( f(0) = 15m + 10 \equiv 3 \pmod{7} \):
     \[
     15m + 10 \equiv 3 \pmod{7} \implies 15m \equiv -7 \pmod{7} \implies 15m \equiv 0 \pmod{7} \implies m \equiv 0 \pmod{7}
     \]
     Thus, \( m = 7n \) for some integer \( n \), and \( f(0) = 15(7n) + 10 = 105n + 10 \).

8. **Smallest non-negative solution**: The smallest non-negative value of \( f(0) \) is when \( n = 0 \):
   \[
   f(0) = 10
   \]

The final answer is \( \boxed{10} \).