We need the least positive value of $a$ such that the quadratic equation has real roots.

1. Rewrite the equation in standard quadratic form

Given: $2x^2 + (a-10)x + \frac{33}{2} = 2a$

Bring all terms to one side: $2x^2 + (a-10)x + \frac{33}{2} - 2a = 0$

So the quadratic is: $2x^2 + (a-10)x + \left(\frac{33}{2} - 2a\right)=0$

Here, $A=2,\quad B=a-10,\quad C=\frac{33}{2}-2a$

2. Condition for real roots

For a quadratic equation to have real roots, its discriminant must satisfy: $D \ge 0$

So, $B^2 - 4AC \ge 0$

Substitute the values: $(a-10)^2 - 4(2)\left(\frac{33}{2}-2a\right) \ge 0$

3. Simplify the discriminant

First compute: $4(2)\left(\frac{33}{2}-2a\right)=8\left(\frac{33}{2}-2a\right)=132-16a$

Thus, $(a-10)^2 - (132-16a) \ge 0$

Expand: $a^2 - 20a + 100 - 132 + 16a \ge 0$ $a^2 - 4a - 32 \ge 0$

Factorize: $a^2 - 4a - 32 = (a-8)(a+4)$

Hence, $(a-8)(a+4) \ge 0$

4. Solve the inequality

Critical points are $a=8$ and $a=-4$.

So, $a \le -4 \quad \text{or} \quad a \ge 8$

We need the least positive value of $a$.

Therefore, $\boxed{8}$

5. Compare with stored answer

Stored correct answer: $8$

Our derived answer is also $8$, so they agree.