To solve the equation

\[
\frac{501}{2015} = \frac{1}{a + \frac{1}{b + \frac{1}{c + \frac{1}{d + \frac{1}{e}}}}}
\]

we start by simplifying the left-hand side:

\[
\frac{501}{2015} = \frac{501 \div 5}{2015 \div 5} = \frac{100.2}{403} \approx \frac{1}{4.003}
\]

We can find fractions equivalent to \( \frac{501}{2015} \) in terms of integers. Using the fact that both numerator and denominator can be simplified:

\[
\frac{501}{2015} = \frac{1}{4}
\]

This leads us to set:

\[
a + \frac{1}{b + \frac{1}{c + \frac{1}{d + \frac{1}{e}}}} = 4
\]

From here we rearrange to find the first component:

\[
\frac{1}{b + \frac{1}{c + \frac{1}{d + \frac{1}{e}}}} = 4 - a
\]

Then we take notice that \( 4 - a \) must be a positive integer for the expression to be valid, meaning \( a \) must be less than 4.

We now consider \( a \) values:

1. **If \( a = 3 \)**:
\[
\frac{1}{b + \frac{1}{c + \frac{1}{d + \frac{1}{e}}}} = 1 \quad \Rightarrow b + \frac{1}{c + \frac{1}{d + \frac{1}{e}}} = 1
\]
Which means \( b = 0 \).

Proceeding with \( b = 0 \):
\[
\frac{1}{c + \frac{1}{d + \frac{1}{e}}} = 1 \quad \Rightarrow c + \frac{1}{d + \frac{1}{e}} = 1
\]
Thus \( c = 0 \).

Continuing:
\[
\frac{1}{d + \frac{1}{e}} = 1 \quad \Rightarrow d + \frac{1}{e} = 1 \quad \Rightarrow d = 0
\]
Finally:
\[
\frac{1}{e} = 1 \quad \Rightarrow e = 1
\]

So, we have:

- \( a = 3 \)
- \( b = 0 \)
- \( c = 0 \)
- \( d = 0 \)
- \( e = 1 \)

Thus, adding these values:

\[
a + b + c + d + e = 3 + 0 + 0 + 0 + 1 = 4
\]

2. **If \( a = 2 \)** or \( a = 1 \)**, similar calculations will lead to non-integer or improper fractions that do not satisfy conditions of \( b, c, d, e \) being integers.

Thus the sum \( a + b + c + d + e \) can only lead to:

\[
\boxed{4}
\]