Four positive integers \(a\), \(b\), \(c\) and \(d\) are such that \[abcd+abc+bcd+cda+dab+ab+bc+cd+da+ac+bd+a+b+c+d=2009.\]

What is the value of \(a+b+c+d\)?

If we add 1 to both sides, the left hand side factorises. We get \[(a+1)(b+1)(c+1)(d+1)=2010.\]

Expressing \(2010\) as a product of primes gives \(2010 = 2 \times 3 \times 5 \times 67\).

As \(2010\) has only four prime factors, these must be \(a+1\), \(b+1\), \(c+1\), and \(d+1\) in some order (since \(a\), \(b\), \(c\) and \(d\) are positive and cannot be zero).

This means that \(a\), \(b\), \(c\), and \(d\) must be \(1\), \(2\), \(4\), and \(66\) in some order.

Thus \[a+b+c+d=1+2+4+66=73.\]