Five, (15 points) A school assigns numbers to the contestants participating in a math competition, with the smallest number being 0001 and the largest number being 2014. No matter which contestant steps forward to calculate the average of the numbers of all other contestants in the school, the average is always an integer. How many contestants can the school have at most?

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Let the school have a total of $n$ participants, whose admission numbers are
$$
1=x_{1}<x_{2}<\cdots<x_{n-1}<x_{n}=2014 .
$$

According to the problem, we have
$$
S_{k}=\frac{x_{1}+x_{2}+\cdots+x_{n}-x_{k}}{n-1}(k=1,2, \cdots, n) \in \mathbf{Z}_{+} .
$$

For any $i, j(1 \leqslant i<j \leqslant n)$, we have
$$
S_{i}-S_{j}=\frac{x_{j}-x_{i}}{n-1} \in \mathbf{Z}_{+} \text {. }
$$

Thus, $x_{j}-x_{i} \geqslant n-1$.
$$
\begin{array}{l}
\text { Hence } x_{n}-x_{1} \\
=\left(x_{n}-x_{n-1}\right)+\left(x_{n-1}-x_{n-2}\right)+\cdots+\left(x_{2}-x_{1}\right) \\
\geqslant(n-1)^{2} \\
\Rightarrow(n-1)^{2} \leqslant x_{n}-x_{1}=2013 \Rightarrow n \leqslant 45 .
\end{array}
$$

Since $\frac{2014-1}{n-1}$ is an integer, it follows that $n-1$ is a divisor of 2013.

Noting that $2013=3 \times 11 \times 61$, the largest divisor not exceeding 45 is 33. Therefore, the maximum value of $n$ is 34, meaning the maximum number of participants is 34.
Such 34 participants' numbers can be realized. For example,
$$
x_{i}=33 i-32(i=1,2, \cdots, 33), x_{34}=2014 \text {. }
$$

Therefore, the maximum number of participants in the competition is 34.