Solution:

Recall that the notation $a \mid b$ means that $a$ divides $b$. We say that a pair of positive integers $(x, y)$ is nice if $x \mid N+y$ and $0<y<x-1$ hold.

The first observation is that, given any positive integer $x'$, among $x'$ consecutive integers there is certainly a multiple of $x'$, so it is always possible to choose (uniquely) an integer $y'$ in the interval $\{0,1, \ldots, x'-1\}$ such that $x' \mid N+y'$. The pair $(x', y')$ is then nice provided that $y'$ is different from $0$ and from $x'-1$. Equivalently, $x'$ is the first member of a nice pair if and only if $x'$ divides neither $N+0=N$ nor $N+(x'-1)$; this second condition is equivalent to the fact that $x'$ does not divide $N-1$.

Let us now consider the nice pair $(x, y)$ for which $x$ is minimal. By the above observation, $x$ can divide neither $N$ nor $N-1$. The conclusion is that there exists some prime power, say $p^{k}$, that divides $x$ but not $N$, and similarly there exists a prime power, say $q^{h}$ (not necessarily different from $p^{k}$), that divides $x$ but not $N-1$. Indeed, if every prime power dividing $x$ also divided $N$, then $x$ itself would divide $N$ by unique factorization, but we have already seen that this cannot happen; the same reasoning applies to $N-1$.

Now observe that, if $q^{h}$ does not divide $N$, then $q^{h}$ divides neither $N$ nor $N-1$, and therefore, by what we have already observed, $q^{h}$ is the first member of a nice pair. On the other hand $q^{h}$ divides $x$, so certainly $q^{h} \leq x$, and since $x$ is minimal, we must have $x=q^{h}$, a power of a prime, as desired. Similarly, if $p^{k}$ does not divide $N-1$, then $x=p^{k}$ and we are done.

We may therefore assume that $p^{k} \mid N-1$ and $q^{h} \mid N$; in this case $p^{k}$ and $q^{h}$ are distinct (recall that $p^{k}$ does not divide $N$) and, since both divide $x$, we get $x \geq 2 p^{k} \geq 2 p$, $x \geq 2 q^{h} \geq 2 q$.

To conclude, we show that in this case one of the numbers $2p$, $2q$ is the first member of a nice pair, and therefore, by minimality, it must coincide with $x$. Indeed, if $N$ is even then $2p$ works: clearly $2p$ cannot divide $N-1$, which is odd, and $p$ (and a fortiori $2p$) cannot divide $N$, because by hypothesis $p$ divides $N-1$ and no prime divides two consecutive numbers. Similarly, if $N$ is odd then $2q$ works: $2q$ (even) does not divide $N$ (odd), and $2q$ does not divide $N-1$ because $q$ divides $N$.