Fourth STARS OF MATHEMATICS COMPETITION, 2010

The necessity of having $x$, $y$, $z$ be pairwise co-prime is proved by, say, assuming $\gcd(x, y) > 1$.
Then $a \mid Nbc + b + c$ becomes $x \mid dNyz + y + z$, and so we must have $\gcd(x, y) \mid z$, absurd, since under this assumption it then follows $\gcd(x, y) \mid \gcd(x, y, z) = 1$.

On the other hand, if the co-primality condition holds, consider the integers
$$xyz - \sum x < 2xyz - \sum x < \dots < (\sum xy) xyz - \sum x.$$
These $\sum xy$ integer numbers will yield different remainders modulo $\sum xy$, since if $ixyz - \sum x \equiv jxyz - \sum x \pmod{\sum xy}$, then also $\sum xy \mid |i - j|xyz$, whence $i = j$, since we have $0 \le |i - j| < \sum xy$ and $\gcd(xyz, \sum xy) = 1$. Therefore there will exist some (unique) $1 \le t \le \sum xy$ such that $\sum xy \mid txyz - \sum x$, i.e. $txyz - \sum x = C \sum xy$ for some positive integer $C$, therefore $txyz = C \sum xy + \sum x$, so $x \mid Cyz + y + z$ et al. We found a suitable value $C$ for the triplet $x$, $y$, $z$ (for similar relations with the ones sought for $a$, $b$, $c$). Then all the other suitable values must be of the form $C' = C + Mxyz$, since we need have $xyz \mid (C' - C) \sum xy$, while $\gcd(xyz, \sum xy) = 1$.

Now the time has come to analyze the last condition. In order to have $a \mid Nbc + b + c$, and the similar others, equivalent to $x \mid dNyz + y + z$ et al., we need have $dN = C + Mxyz$ for some non-negative integer $M$. Denote $e = \gcd(d, xyz)$; then $e \mid d$, so $e \mid C + Mxyz$. But then we also must have $e \mid xyz \mid (C + Mxyz) \sum xy + \sum x$, hence $e \mid \sum x$.
Conversely, if $e \mid \sum x$, then $e \mid xyz \mid (C + Mxyz) \sum xy + \sum x$, so $e \mid C \sum xy$. Since clearly $\gcd(e, \sum xy) = 1$, this means $e \mid C$. Therefore we need have $\frac{d}{e}N = \frac{C}{e} + M\frac{xyz}{e}$, and since clearly $\gcd(\frac{d}{e}, \frac{xyz}{e}) = 1$, take $M \equiv -\frac{C}{e}(\frac{xyz}{e})^{-1} \pmod{d/e}$, wherefore $\frac{d}{e}$ divides $\frac{C}{e} + M\frac{xyz}{e}$. Take now $N = \frac{C + Mxyz}{d}$ (of course, $N' = N + Mabc$ also works).