Suppose that $D$ is a digit and $k$ is a positive integer. Then $D_{(k)}=\underbrace{D D \cdots D D}_{k \text { times }}=D \cdot \underbrace{11 \cdots 11}_{k \text { times }}=D \cdot \frac{1}{9} \cdot \underbrace{99 \cdots 99}_{k \text { times }}=D \cdot \frac{1}{9} \cdot(\underbrace{00 \cdots 00}_{k \text { times }}-1)=D \cdot \frac{1}{9} \cdot\left(10^{k}-1\right)$. Therefore, the following equations are equivalent: $P_{(2k)}-Q_{(k)}=\left(R_{(k)}\right)^{2}$, $P \cdot \frac{1}{9} \cdot\left(10^{2k}-1\right)-Q \cdot \frac{1}{9} \cdot\left(10^{k}-1\right)=\left(R \cdot \frac{1}{9} \cdot\left(10^{k}-1\right)\right)^{2}$, $P \cdot \frac{1}{9} \cdot\left(10^{2k}-1\right)-Q \cdot \frac{1}{9} \cdot\left(10^{k}-1\right)=R^{2} \cdot \frac{1}{81} \cdot\left(10^{k}-1\right)^{2}$, $9P \cdot\left(10^{2k}-1\right)-9Q \cdot\left(10^{k}-1\right)=R^{2} \cdot\left(10^{k}-1\right)^{2}$, $9P \cdot\left(10^{k}-1\right)\left(10^{k}+1\right)-9Q \cdot\left(10^{k}-1\right)=R^{2} \cdot\left(10^{k}-1\right)$ (since $10^{k}-1 \neq 0$), $9P \cdot 10^{k}+9K-9Q=R^{2} \cdot 10^{k}-R^{2}$, $9P-9Q+R^{2}=10^{k}\left(R^{2}-9P\right)$. We consider three cases: $3 \leq k \leq 2018, k=1$, and $k=2$. Case 1: $3 \leq k \leq 2018$. Suppose that $R^{2}-9P \neq 0$. Since $k \geq 3$, then $10^{k}\left(R^{2}-9P\right)>1000$ if $R^{2}-9P>0$ and $10^{k}\left(R^{2}-9P\right)<-1000$ if $R^{2}-9P<0$. Since $P, Q, R$ are digits, then $9P-9Q+R^{2}$ is at most $9(9)-9(0)+9^{2}=162$ and $9P-9Q+R^{2}$ is at least $9(0)-9(9)+0^{2}=-81$. This means that if $R^{2}-9P \neq 0$, we cannot have $9P-9Q+R^{2}=10^{k}\left(R^{2}-9P\right)$ since the possible values do not overlap. So if $3 \leq k \leq 2018$, we must have $R^{2}-9P=0$ and so $9P-9Q+R^{2}=0$. If $R^{2}=9P$, then $R^{2}$ is a multiple of 3 and so $R$ is a multiple of 3. Since $R$ is a positive digit, then $R=3$ or $R=6$ or $R=9$. If $R=3$, then $9P=R^{2}=9$ and so $P=1$. Since $9P-9Q+R^{2}=0$, then $9Q=9(1)+9=18$ and so $Q=2$. If $R=6$, then $9P=R^{2}=36$ and so $P=4$. Since $9P-9Q+R^{2}=0$, then $9Q=9(4)+36=72$ and so $Q=8$. If $R=9$, then $9P=R^{2}=81$ and so $P=9$. Since $9P-9Q+R^{2}=0$, then $9Q=9(9)+81=162$ and so $Q$ cannot be a digit. Therefore, in the case where $3 \leq k \leq 2018$, we obtain the quadruples $(P, Q, R, k)=(1,2,3, k)$ and $(P, Q, R, k)=(4,8,9, k)$. Since there are $2018-3+1=2016$ possible values of $k$, then we have $2 \cdot 2016=4032$ quadruples so far. Case 2: $k=1$. Here, the equation $9P-9Q+R^{2}=10^{k}\left(R^{2}-9P\right)$ becomes $9P-9Q+R^{2}=10R^{2}-90P$ or $99P=9R^{2}+9Q$ or $11P=R^{2}+Q$. For each possible value of $P$ from 1 to 9, we determine the possible values of $Q$ and $R$ by looking for perfect squares that are at most 9 less than $11P$. $P=1$: Here, $11P=11$ which is close to squares 4 and 9. We obtain $(R, Q)=(2,7),(3,2)$. $P=2$: Here, $11P=22$ which is close to the square 16. We obtain $(R, Q)=(4,6)$. $P=3$: Here, $11P=33$ which is close to the square 25. We obtain $(R, Q)=(5,8)$. $P=4$: Here, $11P=44$ which is close to the square 36. We obtain $(R, Q)=(6,8)$. $P=5$: Here, $11P=55$ which is close to the square 49. We obtain $(R, Q)=(7,6)$. $P=6$: Here, $11P=66$ which is close to the square 64. We obtain $(R, Q)=(8,2)$. $P=7$: There are no perfect squares between 68 and 76, inclusive. $P=8$: Here, $11P=88$ which is close to the square 81. We obtain $(R, Q)=(9,7)$. $P=9$: There are no perfect squares between 90 and 98, inclusive. Since $k=1$ in each of these cases, we obtain an additional 8 quadruples. Case 3: $k=2$. Here, the equation $9P-9Q+R^{2}=10^{k}\left(R^{2}-9P\right)$ becomes $9P-9Q+R^{2}=100R^{2}-900P$ or $909P=99R^{2}+9Q$ or $101P=11R^{2}+Q$. As $P$ ranges from 1 to 9, the possible values of $101P$ are 101, 202, 303, 404, 505, 606, 707, 808, 909. As $R$ ranges from 1 to 9, the possible values of $11R^{2}$ are $11,44,99,176,275,396,539,704,891$. The pairs of integers in the first and second lists that differ by at most 9 are (i) 101 and 99 (which give $(P, Q, R)=(1,2,3)$), (ii) 404 and 396 (which give $(P, Q, R)=(4,8,6)$), and (iii) 707 and 704 (which give $(P, Q, R)=(7,3,8)$). Since $k=2$ in each of these cases, we obtain an additional 3 quadruples. In total, there are thus $N=4032+8+3=4043$ quadruples. The sum of the digits of $N$ is $4+0+4+3=11$.