This is a famous problem, sometimes called the coin problem of Frobenius.
If $n=2$ the answer to your question is known to be $N = t_1 t_2 - (t_1 + t_2) + 1$. A proof of this can be found in the answer to this recent question.
For three or more integers, there is no known closed-form solution for $N$. There are some bounds on the values of $N$ in the $n = 3$ case, as well as some algorithms for determining $N$. For more information and references, see the Wikipedia and MathWorld pages on the Frobenius coin problem.
Basically, the problem is considered solved when $n = 2$, partially solved (because of the bounds and algorithms) when $n = 3$, and unsolved when $n \geq 4$.
Update: Guy's Unsolved Problems in Number Theory says, "The case $n = 3$ was first solved explicitly by Selmer and Beyer, using a continued fraction algorithm." So I guess the $n=3$ case has been solved. I suppose you have would have to dig up their paper (it's in the MathWorld references) to see their solution.
Lemma. Let $G$ be a group, written multiplicatively, and let $H$ and $K$ be two subgroups. If $HK = \{hk\mid h\in H, k\in K\}$, then
$$|HK||H\cap K| = |H||K|$$in the sense of cardinalities.
Proof. Consider the map $H\times K\to HK$ given by $(h,k)\mapsto hk$. I claim that the map is exactly $|H\cap K|$ to $1$. Indeed, if $hk=h'k'$, then $h'^{-1}h = k'k^{-1}\in H\cap K$, so there exists $u\in H\cap K$, namely $u=h'^{-1}h$ such that $h=h'u$ and $k=u^{-1}k'$. Thus, $(h,k) = (h'u,u^{-1}k')$ maps to the same thing as $(h',k')$. Conversely, given $v\in H\cap K$, we have that $(h'v,v^{-1}k')\in H\times K$ maps to the same thing as $(h',k')$.
Thus, each element of $HK$ corresponds to precisely $|H\cap K|$ elements of $H\times K$. Thus, $|HK||H\cap K| = |H\times K| = |H||K|$, as claimed. $\Box$
Let $a$ and $b$ be integers, and consider $\mathbb{Z}/\langle ab\rangle$. This is a group with $|ab|$ elements. This group contains subgroups generated by $\gcd(a,b)$, by $a$, by $b$, and by $\mathrm{lcm}(a,b)$. $\gcd(a,b)$ generates the largest subgroup containing both $a$ and $b$; i.e., $\langle \gcd(a,b)\rangle = \langle a\rangle + \langle b\rangle$; while $\mathrm{lcm}(a,b)$ generates the smallest subgroup contained in both $\langle a\rangle$ and $\langle b\rangle$, i.e., $\langle \mathrm{lcm}(a,b)\rangle = \langle a\rangle\cap\langle b\rangle$. By the Lemma (with addition, since we are working in an additive group), we have:
$$|\langle a\rangle+\langle b\rangle| |\langle a\rangle\cap\langle b\rangle| = |\langle a\rangle||\langle b\rangle|$$
Now, the subgroup generated by $\gcd(a,b)$ has $\frac{|ab|}{\gcd(a,b)}$ elements; the subgroup generated by $\mathrm{lcm}(a,b)$ has $\frac{|ab|}{\mathrm{lcm}(a,b)}$ elements; that generated by $a$ has $\frac{|ab|}{|a|}$ elements, that generated by $b$ has $\frac{|ab|}{|b|}$ elements. Plugging all of that in it becomes
$$\gcd(a,b)\mathrm{lcm}(a,b) = |a||b|$$
which yields the desired equality.
$\Box$
Best Answer
It's an equivalence. That is, we have an if and only if.
For the "converse", recall that we have that by Bezout's identity, $(a,b)$ can be written as a linear combination of $a$ and $b$ (this involves the Euclidean algorithm). So if $d$ is the smallest number that has this property, we have $d\le (a,b)$. But of course $d\ge (a,b)$, since any number dividing $a$ and $b$ divides $d$ (by the fact that $d$ is a linear combination of $a$ and $b$). So $d=(a,b)$.