CRT becomes trivial via the innate Arithmetic Progression (A.P.) structure. Note
$$\begin{align} &x\equiv 1\!\!\pmod{\!13}\\ &x\equiv 2\!\!\pmod{\!15}\\ &x\equiv 3\!\!\pmod{\!17}\\[.2em] {\rm i.e.}\ \ \ &x\equiv 1\!+\!k\!\!\!\pmod{\!13\!+\!2k},\ \ k=0,1,2\end{align}$$
with progressions: $\ 1,2,3 = 1\!+\!k,\ $ & $\,\ 13,15,17 = 13\!+\!2k.\,$ Hence
$\!\!\bmod \color{#c00}{13\!+\!2k}\!:\,\ x\equiv 1\!+\!k\overset{\small (2,13+2k)=1}\iff 2x \equiv 2\!+\!\color{#c00}{2k}\equiv 2\color{#c00}{-13}\equiv -11\iff 2x\!+\!11\equiv 0$
So $\ 13,15,17\mid 2x\!+\!11 \!\iff\! n\!=\!13(15)17\mid 2x\!+\!11,\,$ by lcm = product for pair-coprimes.
So $\bmod n\!:\,\ 2x \equiv -11\equiv n\!-\!11\iff x \equiv (n\!-\!11)/2\equiv \bbox[5px,border:1px solid #c00]{\!\!1652}\:$ by $\,n=3315\,$ is odd.
Hence $\ x = 1652 + 3315\,k,\, $ so $\,x = 1652\,$ is clearly the smallest positive solution.
Remark $\ $ If modular fractions are known then more clearly we have by CCRT
$$ x\equiv \dfrac{-11}2\!\!\! \pmod{\!13,15,17} \iff x\equiv \dfrac{-11}2\!\!\! \pmod {\!13\cdot 15\cdot 17}\qquad\qquad$$
More generally the same method shows that if $\,(a,b) = 1\,$ then
$\bbox[8px,border:1px solid #c00]{{\bf Theorem} \ \ \left\{\:\!x\equiv d\!+\!ck\pmod{\!b\!+\!ak}\:\!\right\}_{k=0}^{m_{\phantom{|}}}\!\!\iff\! x\equiv \dfrac{ad\!-\!bc}a\pmod{\!{\rm lcm}\{b\!+\!ak\}_{k=0}^{m_{\phantom{|}}}}} $
Proof $ $ by $\,(a,b\!+\!ak)=(a,b)=1,\,$ LHS $\!\overset{\times\ a}\iff\!\bmod \color{#c00}{b\!+\!ak}\!:\ ax\equiv ad\!+\!c(\color{#c00}{ak})\equiv ad\!\color{#c00}{-b}c\!$ $\!\iff\! ax\!-\!(ad\!-\!bc)\,$ is divisible by all moduli $\!\iff\!$ it is divisible by their lcm $n\, $ (since $\,a\,$ is coprime to each modulus $\,n_k\,$ it is invertible $\!\bmod n_k\,$ so it is invertible mod their lcm $ n)$.
OP is special case $\ a,b,c,d = 2,13,1,1\,$ so $\ x\equiv \dfrac{2(1)\!-\!13(1)}2\equiv\dfrac{-11}2\!\pmod{\!17(16)13}$
See this answer for how to choose the residues in A.P. when only the moduli are in A.P.
Best Answer
We want to assume, as the Comments note, that the GCD of values $n_i$ is 1, as otherwise there is no upper bound on numbers that cannot be represented.
Solving this, the Frobenius coin problem, has previously been discussed here.
Edit: Technically this Question differs from the Coin Problem by requiring the count $a_i$ of each denomination $n_i$ to be "positive", where these coefficients are allowed to be zero in the Coin Problem. Whether or not this was a miswording of the Question, the only effect is to add $n_1 + n_2 + \ldots$ to each possible sum, so the set of representable values is simply shifted upward by this amount.
For general cardinality of $\{n_i\}$, stipulated to be finite, the problem is NP-hard as noted in an on-line paper by D. Beihoffer et al that discusses speed of algorithms. However for fixed number of integers $n_i > 0$ there are algorithms of polynomial complexity in size of input (logarithms of $n_i$'s).