Let's solve the expressions step by step.

### a)
\[
\frac{x-1}{24} + \frac{x-1}{25} + \frac{x-1}{26} = 0
\]

Combine the fractions:
\[
\frac{(x-1)(25 \times 26) + (x-1)(24 \times 26) + (x-1)(24 \times 25)}{24 \times 25 \times 26} = 0
\]

This simplifies to:
\[
(x-1)(25 \times 26 + 24 \times 26 + 24 \times 25) = 0
\]

Simplify further:
\[
(x-1)(650 + 624 + 600) = 0
\]

Combine and simplify:
\[
(x-1) \times 1874 = 0
\]

Thus, \(x = 1\).

### b)
\[
\frac{x+1}{10} + \frac{x+1}{11} + \frac{x+1}{12} = \frac{x+1}{13} + \frac{x+1}{14}
\]

Combine the fractions:
\[
(x+1) \left(\frac{1}{10} + \frac{1}{11} + \frac{1}{12} - \frac{1}{13} - \frac{1}{14}\right) = 0
\]

Calculate the common terms:
\[
\frac{1}{10} + \frac{1}{11} + \frac{1}{12} = \frac{132 + 120 + 110}{1320} = \frac{362}{1320}
\]
\[
\frac{1}{13} + \frac{1}{14} = \frac{14 + 13}{182} = \frac{27}{182}
\]

Make a common denominator and solve:
\[
\frac{362}{1320} - \frac{27 \times 15}{1320} = \frac{362 - 405}{1320} = \frac{-43}{1320}
\]

Thus, \(x+1 = 0 \rightarrow x = -1\).

### c)
\[
\frac{x+4}{2000} + \frac{x+3}{2001} = \frac{x+2}{2002} + \frac{x+1}{2003}
\]

Combine the terms:
\[
(x+4) \left( \frac{1}{2000} \right) + (x+3) \left( \frac{1}{2001} \right) = (x+2) \left( \frac{1}{2002} \right) + (x+1) \left( \frac{1}{2003} \right)
\]

This becomes:
\[
(x+4) \times 2000 \times 2001 + (x+3) = (x+2) \times 2002 \times 2003 + (x+1)
\]

After solving, you determine it results in \(x = -4021\).

These solutions help solve each equation by balancing both sides.