The natural number called $N$ is red iff $(1) \;\; N$ has eight positive divisors $1=d_1 < d_2 < \cdots < d_8=N$ and $(2) \;\; d_5 = 3d_3 - 4$. Assume $N < 2014$ is a red number. According to condition (1) we have the following two cases to consider: Case 1: $N = pqr$, where $p < q < r $ are three primes. Then $(d_1,d_2,d_3) = (1,p,q)$. Assume $r<pq$. Then $(d_4,d_5)= (r,pq)$, yielding $pq = 3q - 4$ by condition (2). Hence $q|4$, which implies $q=2$, contradicting $q > p \geq 2$. Next assume $r>pq$. Then $(d_4,d_5) = (pq,r)$, which means $r = 3q - 4 > pq$ by condition (2). Therefore $p<3$, i.e. $p=2$, which give us $(3) \;\; N = pqr = 2q(3q - 4)$. Now $N < 2014$ means $q(3q - 4) < 1007$, yielding $q < 19$. Hence $(q,r) = (q,3q-4) = (3, 5), (5,11), (7, 17), (11,29), (13,35), (17,47)$. Thus we have the following red numbers: $N = 2 \cdot 3 \cdot 5 = 30$ $N = 2 \cdot 5 \cdot 11 = 110$ $N = 2 \cdot 7 \cdot 17 = 238$ $N = 2 \cdot 11 \cdot 29 = 638$ $N = 2 \cdot 17 \cdot 47 = 1598$ Case 2: $N = p^3q$, where $p$ and $q$ are primes. First assume $q<p$. Then $(d_1,d_2,d_3,d_4,d_5) = (1, q, p, pq, p^2)$, yielding $p^2 = 3p - 4$ by condition (2). Hence $p | 4$, which means $p=2$, which give us the contradiction $0 = p^2 - 3p + 4 = 2^2 - 3 \cdot 2 + 4 = 2$. Assume $p < q < p^2$. Then $(d_1,d_2,d_3,d_4,d_5) = (1, p, q, p^2, pq)$, which give us $pq = 3q - 4$ by condition (2). Consequently $q|4$, meaning $q=2$, which contradicts $q > p \geq 2$. Next assume $p^2 < q < p^3$. Then $(d_1,d_2,d_3,d_4,d_5) = (1, p, p^2, q, p^3)$, yielding $p^3 = 3p^2 - 4$ by condition (2). Hence $p=2$ and $d_3 = p^2 = 2^2 = 4 < d_3 = q < d_4 = p^3 = 2^3 = 8$. Therefore $q=5$ or $q=7$, which means $N=2^3 \cdot 5 = 40$ and $N=2^3 \cdot 7 = 56$ are red numbers. Finally assume $q > p^3$. Then $(d_1,d_2,d_3,d_4,d_5) = (1, p, p^2, p^3, q)$, which yields $q = 3p^2 - 4$ by condition (2). From $q> p^3$ we obtain $p^2(p - 3) < -4$, implying the contradiction $p<2$. Conclusion: There are seven red numbers smaller than 2014.