From floor to ceiling

Find a positive integer $n$ such that $\left\lfloor \frac {20n}{13} \right\rfloor + \left\lceil \frac {13n}{20} \right\rceil = 2013$.

This problem is proposed by Ahaan Rungta.

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Details and assumptions:

• The function $\lfloor x \rfloor: \mathbb{R} \rightarrow \mathbb{Z}$ refers to the greatest integer smaller than or equal to $x$. For example $\lfloor 2.3 \rfloor = 2$ and $\lfloor -5 \rfloor = -5$.

• The function $\lceil x \rceil : \mathbb{R} \rightarrow \mathbb{Z}$ refers to the smallest integer that is greater than or equal to to $x$. For example, $\lceil 2.3 \rceil = 3$ and $\lceil -5 \rceil = -5$.