Irish Mathematical Olympiad

The answer is $2015$. To prove this, first, observe (using Euclid's algorithm, for example), that
$$15 \cdot 17 - 2 \cdot 127 = 1 \qquad (1)$$
Next, note that
$$\begin{align*} 2016 &= (17 - 1)(127 - 1) = 16 \cdot 127 - 17 + 1 \\ &= 16 \cdot 127 - 17 + 15 \cdot 17 - 2 \cdot 127 \tag{2} \\ &= 14 \cdot 127 + 14 \cdot 17 \end{align*}$$
We now claim: If $z > 2016$ is an integer, then $z = 127a + 17b$, for some non-negative integers $a, b$.
To prove the claim, we note that by (1), $z = 127c + 17d$, for some integers $c, d$, and then $z = 127(c + 17y) + 17(d - 127y)$, for any integer $y$. So we can write
$$z = 127a + 17b, \text{ with integers } a, b \text{ with } a \ge 0.$$
Choose such a representation with $b$ greatest possible. If $b \ge 0$, the Claim is established. Suppose, for the sake of contradiction, that $b < 0$. Then $z = 127(a - 17) + 17(b + 127)$, and, thus, by our choice of $b$, $a - 17 < 0$, and thus $z = 127a + 17b \le 127 \cdot 16 - 17 < 2016$, contradicting our hypotheses. So the Claim is proved.
Combining this with (2), we see that every integer $w \ge 2016$ is expressible as $w = 17x + 127y$, for some non-negative integers $x, y$.
Finally, suppose that $2015 = 16 \cdot 127 - 17 = 17u + 127v$, for some non-negative integers $u, v$. Then, using subtraction, $127(16 - v) = 17(u + 1)$. This shows that the positive number $u + 1$ is divisible by $127$ and so $u \ge 126$, implying that $2015 \ge 17 \cdot 126$, which is false. Hence $2015$ is the integer sought.