The fact $a_1,a_2,\ldots,a_n$ are the sides of a polygon translates into $0< 2a_k < p$ for all $1\leq k \leq n$. We then have $2a_k^2 < 2pa_k - pa_k$, or $pa_k < 2a_k(p - a_k)$, thus $\dfrac {a_k} {p-a_k} < \dfrac {2a_k} {p}$. Finally, $\sum_{k=1}^n \dfrac {a_k} {p-a_k} < \sum_{k=1}^n \dfrac {2a_k} {p} = \dfrac {2p} {p} = 2$. Notice that, whereas equality cannot be reached, the upper bound of $2$ is the best possible, since for $a_1=a_2= 1$ and $a_3=\cdots = a_n = \varepsilon > 0$, we have $p = 2 + (n-2)\varepsilon$ and $\sum_{k=1}^n \dfrac {a_k} {p-a_k} = \dfrac {2} {1 + (n-2)\varepsilon} + \dfrac {(n-2)\varepsilon} {2 + (n-3)\varepsilon}$, and so $\lim_{\varepsilon \to 0} \sum_{k=1}^n \dfrac {a_k} {p-a_k} = 2$. This quite a well-known inequality; I wonder how and why it was selected for a competition, in the year 2013 ... In the different direction, for just positive real numbers $a_1,a_2,\ldots,a_n$ we have $\sum_{k=1}^n \dfrac {a_k} {p-a_k} = p\sum_{k=1}^n \dfrac {1} {p-a_k} - n =$ $ \dfrac {1} {n-1} \left (\sum_{k=1}^n (p-a_k) \right )\left (\sum_{k=1}^n \dfrac {1} {p-a_k}\right ) - n \geq \dfrac {n^2} {n-1} - n =$ $ \dfrac {n} {n-1}$, with equality for $a_1=a_2=\cdots = a_n$. The case $n=3$ is Nesbitt's inequality.

A different solution.. Observe - \[\frac{a_1}{p-a_1}+\frac{a_2}{p-a_2}+\cdots + \frac{a_n}{p-a_n} < 2 \] \[\implies \frac{a_1}{p-a_1}+1 \frac{a_2}{p-a_2}+1 \cdots + \frac{a_n}{p-a_n}+1 < 2+n\] \[\implies \frac{p}{p-a_1}+\frac{p}{p-a_2}+\cdots + \frac{p}{p-a_n} < 2+n \] \[\implies \frac{1}{p-a_1}+\frac{1}{p-a_2}+\cdots + \frac{1}{p-a_n} < \frac{2}{p}+\frac{n}{p} \] Denote \[\frac{1}{p-a_1}+\frac{1}{p-a_2}+\cdots + \frac{1}{p-a_n} = \chi \] For any $0<i\leq{n}$, $\frac{1}{p-a_i}>\frac{1}{p}$ $\therefore$ $\chi>\frac{n}{p}$ But, $\frac{1}{p-a_i}-\frac{1}{p}=\frac{a_i}{p(p-a_k)}<\frac{a_i}{p}$ $\implies$ $\chi<\frac{a_1+a_2+\cdots+a_n}{p}$ $\implies$ $\chi<1$. Since, we already proved that $\chi<\frac{2}{p}+\frac{n}{p}$ It suffices to show that $\frac{2}{p}<1$. Which is trivial.. Therefore we may conclude that the given statement is true.. Q.E.D. $\square$

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Observe that $\frac{a_i}{p-a_{i}}<\frac{2a_i}{p}$ since $p>2a_i$. Sum cyclically to finish. Credits: SiddhantAg