= Solution
The <Lagrange theorem for polynomial congruences> says that if $p$ is prime and $f\in\mathbb Z[X]$ has degree $d$ with at least one coefficient not divisible by $p$, then
$$
f(x)\equiv0\pmod p
$$
has at most $d$ incongruent solutions modulo $p$.
The <Chinese remainder theorem> states that for pairwise coprime $m_1,\ldots,m_r$, every system
$$
x\equiv a_i\pmod{m_i}
\qquad(1\leq i\leq r)
$$
has exactly one solution modulo $M=\prod_i m_i$. For two moduli, choose $u,v$ with $um+vn=1$ by <Bezout identity>. Then
$$
x=a\,vn+b\,um
$$
is congruent to $a$ modulo $m$ and to $b$ modulo $n$. If $x,y$ are two solutions, both $m$ and $n$ divide $x-y$; coprimality makes $mn$ divide $x-y$. This proves existence and uniqueness for two factors, and induction proves the general statement.
Now
$$
1729=7\cdot13\cdot19
$$
and $12^3+1=1729$, so $x=12$ is a solution. For $1\leq x<12$, one has $0<x^3+1<1729$, so no such positive integer can satisfy the congruence. Hence the smallest is
$$
\boxed{x=12}.
$$
Modulo $7,13,19$, the roots are respectively
$$
\{3,5,6\},\qquad
\{4,10,12\},\qquad
\{8,12,18\}.
$$
Each list has three elements, and the Chinese remainder theorem combines the choices independently. Therefore the number of solutions with $1\leq x\leq1729$ is
$$
\boxed{3^3=27}.
$$
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