Here is a simple proof. By the theorem you quote in your comment to Ross's answer, we know that all of the solutions of $\rm\:N\ =\ 4\ X + 7\ Y\:$ arise from adding or subtracting multiples of $\rm\:(-7,4)\:$ to any particular solution $\rm\:(X,Y).\:$ Thus we may normalize any representation $\rm\ N\ =\ 4\ X + 7\ Y\ $ so that $\rm\ 0 \le X \le 6,\: $ by adding some integral multiple of $\rm\ (-7,4)\ $ to $\rm\:(X,Y),\:$ i.e. by choosing $\rm\,K\,$ so that $\rm\:(X,Y)+K(-7,4)\, =\, (X\!-\!7K,\,Y\!+\!4K)\:$ has $\rm\:0\le X\!-\!7K \le 6.\:$ Then we observe
Lemma $\rm\ \ N \,=\, 4\, X\! + 7\, Y\ $ for some integers $\rm\ X,Y \ge 0\, $
$\iff$ its normalization has $\rm\: Y \ge 0\:$.
Proof $\rm\ \ (\Rightarrow)\ \ $ If $\rm\ X,Y \ge 0\ $ then normalization adds $\rm\:(-7,4)\:$ zero or more times (in order to get $\rm\:0 \le X \le 6),\ $ and this preserves the condition $\rm\: Y \ge 0\:.\ $ $\ (\Leftarrow)\ \ $ Conversely if the normalization has $\rm\: Y < 0\:,\ $ then $\rm\:N\:$ has no
representation with $\rm\ X, Y \ge 0\:,\: $ because to shift $\rm\: Y > 0\: $
requires adding $\rm\ (-7,4)\ $ at least once, which (since $\rm\:0\le X\le 6),\:$ shifts $\rm\: X < 0\:.\ $ QED
Finally, since $\rm\ 4\,X\! + 7\, Y\ $ is increasing in both $\rm\: X,Y\:,\ $
it is clear that the largest non-representable number $\rm\: N\:$ has normalization
$\rm\: (X,Y)\ =\ (6,-1)\:,\: $ corresponding to $\rm\ N\ =\ 4\cdot 6 - 1\cdot 7\, =\, 17.$
Precisely the same argument works also for the general case. This classic problem is known by a great variety of aliases, e.g. postage stamp problem, Sylvester/Frobenius problem, Diophantine problem of Frobenius, Frobenius conductor, money changing, coin changing, change making problems. See also: h-basis and asymptotic bases in additive number theory, integer programming algorithms and Gomory cuts, knapsack problems and greedy algorithms, etc.