Let $f: \mathbb{R} \to \mathbb{R}$ be continuous and satisfy, for every $a < b$,
$$f\left(\frac{a+b}{2}\right) = \frac{m_{ab} + M_{ab}}{2}$$
where $m_{ab} = \min_{x \in [a, b]} f(x)$ and $M_{ab} = \max_{x \in [a, b]} f(x)$.

Let us fix $a < b$ and consider the function $f$ on $[a, b]$.

Since $f$ is continuous on $[a, b]$, it attains its minimum and maximum at some points $x_1, x_2 \in [a, b]$:
$$m_{ab} = f(x_1), \quad M_{ab} = f(x_2)$$

Let us consider the value at the midpoint $m = \frac{a+b}{2}$:
$$f(m) = \frac{f(x_1) + f(x_2)}{2}$$

But $f(m)$ must also be between $m_{ab}$ and $M_{ab}$, i.e., $f(m) \in [m_{ab}, M_{ab}]$.

Suppose $f$ is not constant on $[a, b]$. Then $m_{ab} < M_{ab}$, and $f(m)$ is strictly between them unless $f(m)$ equals one of the endpoints. But $f(m)$ is the average of the minimum and maximum, so unless $f$ is constant, $f(m)$ is strictly between $m_{ab}$ and $M_{ab}$.

Now, let us consider the following:

Let $f$ be strictly increasing. Then $m_{ab} = f(a)$, $M_{ab} = f(b)$, and $f(m) = \frac{f(a) + f(b)}{2}$.
But for a strictly increasing continuous function, $f(m)$ is strictly less than $f(b)$ and strictly greater than $f(a)$, unless $f$ is linear.

Let us try $f(x) = cx + d$ (affine function).

Then $f(a) = ca + d$, $f(b) = cb + d$, $f(m) = c\frac{a+b}{2} + d = \frac{ca + cb}{2} + d = \frac{f(a) + f(b)}{2}$.

On $[a, b]$, since $f$ is affine, the minimum and maximum are at the endpoints:
If $c > 0$, $m_{ab} = f(a)$, $M_{ab} = f(b)$.
If $c < 0$, $m_{ab} = f(b)$, $M_{ab} = f(a)$.
If $c = 0$, $f$ is constant.

In both cases, $f(m) = \frac{f(a) + f(b)}{2}$, and $m_{ab} + M_{ab} = f(a) + f(b)$, so the condition is satisfied.

Now, suppose $f$ is not affine. For example, suppose $f$ is quadratic: $f(x) = x^2$.
On $[a, b]$, the minimum is at $a$ or $b$ (if $0 \notin [a, b]$), or at $0$ (if $0 \in [a, b]$).
Suppose $a < 0 < b$.
Then $m_{ab} = 0$, $M_{ab} = \max\{a^2, b^2\}$.
But $f(m) = (\frac{a+b}{2})^2$.

Is it always true that $f(m) = \frac{m_{ab} + M_{ab}}{2}$? For $a = -1$, $b = 1$, $m = 0$, $f(m) = 0$, $m_{ab} = 0$, $M_{ab} = 1$, so $\frac{m_{ab} + M_{ab}}{2} = \frac{1}{2}$. But $f(m) = 0 \neq \frac{1}{2}$.

Therefore, $f(x) = x^2$ does not satisfy the condition.

Suppose $f$ is constant: $f(x) = d$.
Then $m_{ab} = d$, $M_{ab} = d$, $f(m) = d$, $\frac{m_{ab} + M_{ab}}{2} = d$.
So the condition is satisfied.

Therefore, the only continuous functions $f: \mathbb{R} \to \mathbb{R}$ satisfying the condition are the affine functions $f(x) = cx + d$.

Let us check for $c < 0$:
Then $m_{ab} = f(b)$, $M_{ab} = f(a)$, $f(m) = \frac{f(a) + f(b)}{2}$, $\frac{m_{ab} + M_{ab}}{2} = \frac{f(a) + f(b)}{2}$.
So the condition is satisfied.

Thus, all continuous functions $f: \mathbb{R} \to \mathbb{R}$ of the form $f(x) = cx + d$ for $c, d \in \mathbb{R}$ satisfy the condition.

Final answer:

All continuous functions $f: \mathbb{R} \to \mathbb{R}$ of the form $f(x) = cx + d$ for $c, d \in \mathbb{R}$.