"When I was 4"

We use the factorisation $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$ .

$\begin{aligned} \dfrac{x^3+y^3}{3} +xy &= 2007\\ x^3+y^3+3xy &= 6021\\ x^3+y^3+(-1)^3 - 3xy(-1) &= 6020\\ (x+y-1)(x^2+y^2+1-xy+x+y) &= 6020\\ (x+y-1)((x+y-1)^2 - 3xy - 3x - 3y) &= 2^2 \times 5 \times 7 \times 43 \end{aligned}$

Since the RHS $\equiv 2 \pmod{3}$ , the LHS must also be $\equiv 2 \pmod{3}$ . Thus, we must have $x+y-1 \equiv 2 \pmod{3}$ . All positive factors of 6020 that are equivalent to $2 \pmod{3}$ are $2$ , $5$ , $14$ , $20$ , $35$ , $86$ , $140$ , $215$ , $602$ , $860$ , $1505$ and $6020$ . By observation, we get that only $x+y-1=20$ satisfies, with $x=18$ , $y=3$ or vice-versa. Thus, $x+y=21$ , but since we're looking for ordered pairs, the answer is $\boxed{42}$ .

Since $6021 =x^3 + y^3 + 3xy = (x+y)^3 - 3xy(x+y) + 3xy$ we see that $x+y$ is divisible by $3$ . But then $3xy = 6021 + 3xy(x+y) - (x+y)^3$ is divisible by $9$ , and so $3$ divides $xy$ . Thus both $x$ and $y$ are divisible by $3$ , and so $x=3X$ , $y=3Y$ where $\begin{aligned} 6021 & = \; 27(X+Y)^3 - 81XY(X+Y) + 27XY \\ 27XY & = \; \frac{27(X+Y)^3 - 6021}{3(X+Y) - 1} \; = \; 9(X+Y)^2 + 3(X+Y) + 1 - \frac{6020}{3(X+Y)-1} \end{aligned}$ so we deduce that $3(X+Y)-1$ divides $6020$ . Possible values of $X+Y$ , and therefore $XY$ , are thus

$\begin{array}{|c|c|c|c|c|c|c|c|} \hline X+Y & XY & X+Y & XY & X+Y & XY & X+Y & XY \\ \hline -1003 & 335225 & -401 & 53556 & -143 & 6801 & -100 & 3323 \\ -57 & 1078 & -23 & 177 & -14 & 69 & -9 & 34 \\ -3 & 25 & -2 & 33 & -1 & 56 & 0 & 223 \\ 1 & -111 & 2 & -43 & 5 & -7 & 7 & 6 \\ 12 & 43 &29 & 281 & 47 & 740 & 72 & 1735 \\ 201 & 13489 & 287 & 27488 & 502 & 84057 & 2007 & 1342906 \\ \hline \end{array}$ Of these, only the pair $X+Y=7$ , $XY= 6$ correspond to integer solutions $X,Y = 1,6$ . Thus the only solutions to the equation are $(x_1,y_1) = (18,3)$ and $(x_2,y_2) = (3,18)$ , which makes the answer $\boxed{42}$ .