There are positive integers $b$ and $c$ such that the polynomial $2x^2 + bx + c$ has two real roots which differ by $30.$ Find the least possible value of $b + c.$
1. **Identify the roots of the polynomial**: The polynomial given is \(2x^2 + bx + c\). The roots of this polynomial can be found using the quadratic formula:
   \[
   x = \frac{-b \pm \sqrt{b^2 - 8c}}{4}
   \]
   Let the roots be \(r_1\) and \(r_2\). According to the problem, the difference between the roots is 30.

2. **Express the difference between the roots**: The difference between the roots \(r_1\) and \(r_2\) is given by:
   \[
   r_1 - r_2 = \frac{\sqrt{b^2 - 8c}}{2}
   \]
   Setting this equal to 30, we get:
   \[
   \frac{\sqrt{b^2 - 8c}}{2} = 30
   \]
   Solving for \(\sqrt{b^2 - 8c}\), we multiply both sides by 2:
   \[
   \sqrt{b^2 - 8c} = 60
   \]
   Squaring both sides, we obtain:
   \[
   b^2 - 8c = 3600
   \]

3. **Solve for \(c\)**: Rearrange the equation to solve for \(c\):
   \[
   8c = b^2 - 3600
   \]
   \[
   c = \frac{b^2 - 3600}{8}
   \]

4. **Ensure \(c\) is a positive integer**: Since \(c\) must be a positive integer, \(b^2 - 3600\) must be a positive multiple of 8. Therefore, \(b^2\) must be a perfect square greater than 3600 and divisible by 8.

5. **Find the smallest \(b\) such that \(b^2\) is a multiple of 8 and greater than 3600**: The smallest perfect square greater than 3600 that is also a multiple of 8 is \(64^2\). Let's verify:
   \[
   64^2 = 4096
   \]
   Substituting \(b = 64\) into the equation for \(c\):
   \[
   c = \frac{64^2 - 3600}{8} = \frac{4096 - 3600}{8} = \frac{496}{8} = 62
   \]

6. **Calculate \(b + c\)**: Now, we find the sum \(b + c\):
   \[
   b + c = 64 + 62 = 126
   \]

Conclusion:
\[
\boxed{126}
\]