I'll write down the solution I gave during the contest. First, it's easy to see that the constant function $f(x)=\frac{t}{b-a}$ verifies the problem. Now we want to prove that every function with the property $(P)$ is constant. Let $x \in [a,b]$ be an arbitrary number. Rewriting, we get $f(x)=\frac{f(\frac{a+x}{2})+f(\frac{x+b}{2})}{2}$. Writing the property for $x:=\frac{a+x}{2}$ and for $x:=\frac{x+b}{2}$ we get $f(\frac{a+x}{2})=\frac{f(\frac{3a+x}{4})+f(\frac{a+x+2b}{4})}{2}$ and $f(\frac{x+b}{2})=\frac{f(\frac{2a+x+b}{4})+f(\frac{x+3b}{4})}{2}$. So we now get $f(x)=\frac{f(\frac{3a+x}{4})+f(\frac{a+x+2b}{4})+f(\frac{2a+x+b}{4})+f(\frac{x+3b}{4})}{4}$. Now the intuition tells us that we could write $(b-a)f(x)$ as some kind of Riemann sum with the norm $\frac{b-a}{2^n}$, this is because $a \leq \frac{a+x}{2} \leq \frac{a+b}{2} \leq \frac{x+b}{2} \leq b $ and $ a \leq \frac{3a+x}{4} \leq \frac{3a+b}{4} \leq \frac{2a+x+b}{4} \leq \frac{a+b}{2} \leq \frac{a+x+2b}{4} \leq \frac{x+3b}{4} \leq b$, so the base cases $n=1$ and $n=2$ both work. We will now prove by induction that $(b-a)f(x)$ can be written as a Riemann sum of norm $\frac{b-a}{2^n}, \forall n\in\mathbb{N}$. I won't write the full solution because it's pretty lengthy, but consider the equidistant division $\Delta_n=(a=x_{0,n}, x_{1,n}=a+\frac{b-a}{2^n},..., x_{k,n}=a+\frac{k(b-a)}{2^n},...,x_{2^n,n}=b)$ and we will prove the induction (intuitively) in the following way : Assume $P(n)$ is true, i.e. $(b-a)f(x)$ can be written as a Riemann sum of norm $\frac{b-a}{2^n}$, and notice now that the "$x_{k,n}$" in $\Delta_n$ "brings" the $k$-th and the $(2^n+k)$-th term of the Riemann sum in $P(n+1)$ when applying the property $f$ has for $x:=x_{k,n}$, i.e. $f(x_{k,n})=\frac{f(x_{k,n+1})+f(x_{2^n+k, n+1})}{2}$. This is indeed true (exercise), and makes the transition from $P(n)$ to $P(n+1)$. Now that we proved the induction, taking the limit as $n \to \infty$, since the Riemann sum converges to the integral and $ \int_{a}^{b} f(x) dx=t$ we get that $(b-a)f(x)=t$, i.e. $f(x)=\frac{t}{b-a}$, which is exactly what we wanted to prove. Since $x$ was arbitrary from $[a,b]$, we're done.