Find the pairs of real numbers $(a,b)$ such that the biggest of the numbers $x=b^2-\frac{a-1}{2}$ and $y=a^2+\frac{b+1}{2}$ is less than or equal to $\frac{7}{16}$
1. We start with the given equations:
   \[
   x = b^2 - \frac{a - 1}{2}
   \]
   \[
   y = a^2 + \frac{b + 1}{2}
   \]
   and the condition:
   \[
   \max(x, y) \leq \frac{7}{16}
   \]

2. To satisfy the condition \(\max(x, y) \leq \frac{7}{16}\), both \(x\) and \(y\) must be less than or equal to \(\frac{7}{16}\). Therefore, we have:
   \[
   x \leq \frac{7}{16}
   \]
   \[
   y \leq \frac{7}{16}
   \]

3. Substitute the expressions for \(x\) and \(y\) into these inequalities:
   \[
   b^2 - \frac{a - 1}{2} \leq \frac{7}{16}
   \]
   \[
   a^2 + \frac{b + 1}{2} \leq \frac{7}{16}
   \]

4. Simplify the first inequality:
   \[
   b^2 - \frac{a - 1}{2} \leq \frac{7}{16}
   \]
   \[
   b^2 - \frac{a}{2} + \frac{1}{2} \leq \frac{7}{16}
   \]
   \[
   b^2 - \frac{a}{2} \leq \frac{7}{16} - \frac{1}{2}
   \]
   \[
   b^2 - \frac{a}{2} \leq \frac{7}{16} - \frac{8}{16}
   \]
   \[
   b^2 - \frac{a}{2} \leq -\frac{1}{16}
   \]

5. Simplify the second inequality:
   \[
   a^2 + \frac{b + 1}{2} \leq \frac{7}{16}
   \]
   \[
   a^2 + \frac{b}{2} + \frac{1}{2} \leq \frac{7}{16}
   \]
   \[
   a^2 + \frac{b}{2} \leq \frac{7}{16} - \frac{1}{2}
   \]
   \[
   a^2 + \frac{b}{2} \leq \frac{7}{16} - \frac{8}{16}
   \]
   \[
   a^2 + \frac{b}{2} \leq -\frac{1}{16}
   \]

6. Combine the two inequalities:
   \[
   b^2 - \frac{a}{2} \leq -\frac{1}{16}
   \]
   \[
   a^2 + \frac{b}{2} \leq -\frac{1}{16}
   \]

7. Notice that both \(b^2\) and \(a^2\) are non-negative, and \(\frac{a}{2}\) and \(\frac{b}{2}\) are real numbers. For the inequalities to hold, the only possibility is that both \(b^2\) and \(a^2\) must be zero, and the linear terms must balance out the negative constant term. This implies:
   \[
   b^2 = 0 \quad \text{and} \quad a^2 = 0
   \]
   \[
   b = 0 \quad \text{and} \quad a = 0
   \]

8. However, substituting \(a = 0\) and \(b = 0\) into the original equations for \(x\) and \(y\) does not satisfy the inequalities. Therefore, we need to re-evaluate the conditions.

9. Reconsider the inequalities:
   \[
   b^2 - \frac{a}{2} \leq -\frac{1}{16}
   \]
   \[
   a^2 + \frac{b}{2} \leq -\frac{1}{16}
   \]

10. The only way for these inequalities to hold is if both \(a\) and \(b\) are such that the quadratic terms are zero and the linear terms balance out the negative constant term. This leads us to:
    \[
    4a - 1 = 0 \quad \text{and} \quad 4b + 1 = 0
    \]
    \[
    a = \frac{1}{4} \quad \text{and} \quad b = -\frac{1}{4}
    \]

11. Verify the solution:
    \[
    x = \left(-\frac{1}{4}\right)^2 - \frac{\frac{1}{4} - 1}{2} = \frac{1}{16} - \frac{-\frac{3}{4}}{2} = \frac{1}{16} + \frac{3}{8} = \frac{1}{16} + \frac{6}{16} = \frac{7}{16}
    \]
    \[
    y = \left(\frac{1}{4}\right)^2 + \frac{-\frac{1}{4} + 1}{2} = \frac{1}{16} + \frac{\frac{3}{4}}{2} = \frac{1}{16} + \frac{3}{8} = \frac{1}{16} + \frac{6}{16} = \frac{7}{16}
    \]

12. Both \(x\) and \(y\) are equal to \(\frac{7}{16}\), satisfying the condition \(\max(x, y) \leq \frac{7}{16}\).

The final answer is \( \boxed{ (a, b) = \left(\frac{1}{4}, -\frac{1}{4}\right) } \)