$34+59+17+118 = 228$ $34 \cdot 59 \cdot 17 \cdot 118 \equiv \ 0 \ \text{mod}\ 2006^{2}$

$2006^{2}= 2^{2}\cdot 17^{2}\cdot 59^{2}$ By a little testing, we find the solution $59+59 \cdot 2+17+17 \cdot 2 = \boxed{228}$. Now, to prove that this is the minimum: First, note that $17^{2}> 228$. So, this means that the two factors of $17$ must be kept in separate terms, as well as the two factors of $59$. Then, since the terms must be distinct, we must distribute the two $2$s, one to a factor of $59$, and the other to a factor of $17$. This gives the solution above. Now, if we were to have that the product is a larger multiple of $2006^{2}$, we would need either to multiply the extra factors into one of the terms in our sum, or add it at the end. Both would increase the sum; so, we have found our minimum. $\Box$