How can I solve $T(n) = \lfloor \frac {n} {2}\rfloor + T(\lceil \frac{n}{2} \rceil)$? I came u with the following equation:
$T(2) = 1$
$T(n) = \lfloor \frac {n} {2}\rfloor + T(\lceil \frac{n}{2} \rceil) | n ∈ \mathbb N$
But I can't find a way to solve it. Is there one? 
I need to proof that the solution is right, so I can't just look at the numerical sequence of the equation.

You can solve this as follows. First calculate the first few values 
\begin{align}
T(3) = 1 + T(2) = 2 \\
T(4) = 2 + T(2) = 3 \\ 
T(5) = 2 + T(3) = 4 \\ 
T(6) = 3 + T(3) = 5
\end{align}
Ok, it might be that T(n) = n - 1. Let's try to prove this by induction. Firstly, we have $T(2) = 1$ so this is obviously true for $n = 2$. Now assume it is true for $n = 2, 3 , \dots k$ and we would now like to prove it for $n= k+1$. 
\begin{equation}
T(k+1) = \left\lfloor \frac{k+1}{2} \right\rfloor + T \left( \left\lceil \frac{k+1}{2}\right\rceil\right).
\end{equation}
Now, we must consider two different cases. Either $k+1$ is even or odd. If $k$ is even the above equation becomes 
\begin{equation}
T(k+1) = \frac{k+1}{2} + T\left( \frac{k+1}{2} \right) = \frac{k+1}{2} + \left( \frac{k+1}{2} - 1 \right) = (k+1) - 1 . 
\end{equation}
Above I used that we have assumed that $T(n) = n-1$ for $n\leq k$ in the second step. Now assume that $k+1$ is odd. We then get
\begin{equation}
T(k+1) = \frac{k+1}{2} - 0.5 + T\left(\frac{k+1}{2} + 0.5 \right) = 
\frac{k+1}{2} - 0.5 + \left(\frac{k+1}{2} + 0.5 -1 \right) = (k+1)-1.
\end{equation}
Hence, by the induction principle, this is true for all $n$. 
