First see that $\gcd(6n+11,6n+14) = 1$, $\gcd(6n+11,20n+19) = 1$ or $53$, and $\gcd(6n+14,20n+19) = 1$ or $83$. If the expression is a rational number, then it must be an integers, because it's algebraic. Hence we have $4$ options: $1^\circ$ $6n+11 = a^2$, $6n+14 = b^2$, $20n+19 = c^2$, here it's obvious contradiction, because $b^2 - a^2 = 3$, so $b = 2$ and $a=1$, but then $n$ can't be an integer. $2^\circ$ $6n+11 = 53a^2$, $6n+14 = b^2$, $20n+19 = 53c^2$, then we have $b^2 - 53a^2 = 3$, so $b^2 \equiv 3 \pmod{53}$, but $\left(\frac{3}{53}\right) = \left(\frac{53}{3}\right)(-1)^{(3-1)(53-1)/4} = \left(\frac{2}{3}\right)\cdot1 = -1$, contradiction. $3^\circ$ $6n+11 = a^2$, $6n+14 = 83b^2$, $20n+19 = 83c^2$, then we have $83b^2 - a^2 = 3$, so $a^2 \equiv -3 \pmod{83}$, but $\left(\frac{-3}{83}\right) = \left(\frac{-1}{83}\right)\left(\frac{83}{3}\right)(-1)^{(3-1)(83-1)/4} = (-1)\cdot\left(\frac{2}{3}\right)\cdot(-1) = -1$, contradiction. $4^\circ$ We are left with $6n+11 = 53a^2$, $6n+14 = 83b^2$, $20n+19 = 53\cdot83c^2$. Now by some calculations we can get $b \equiv 4 \pmod {53}$ and $a \equiv 5 \pmod {83}.$ Putting $a = 5$ and $b = 4$ we get that $n = 219$ and $c = 1$ with $20n+19 = 4399 = 58\cdot 83$. We know that $a$, $b$ are the smallest then, so it will be the smallest $n = 219$ such that the product is a square.

Impossible solution for a children havimg 13-14 years...

TheMathBob wrote: First see that $\gcd(6n+11,6n+14) = 1$, $\gcd(6n+11,20n+19) = 1$ or $53$, and $\gcd(6n+14,20n+19) = 1$ or $83$. If the expression is a rational number, then it must be an integers, because it's algebraic. Hence we have $4$ options: $1^\circ$ $6n+11 = a^2$, $6n+14 = b^2$, $20n+19 = c^2$, here it's obvious contradiction, because $b^2 - a^2 = 3$, so $b = 2$ and $a=1$, but then $n$ can't be an integer. $2^\circ$ $6n+11 = 53a^2$, $6n+14 = b^2$, $20n+19 = 53c^2$, then we have $b^2 - 53a^2 = 3$, so $b^2 \equiv 3 \pmod{53}$, but $\left(\frac{3}{53}\right) = \left(\frac{53}{3}\right)(-1)^{(3-1)(53-1)/4} = \left(\frac{2}{3}\right)\cdot1 = -1$, contradiction. $3^\circ$ $6n+11 = a^2$, $6n+14 = 83b^2$, $20n+19 = 83c^2$, then we have $83b^2 - a^2 = 3$, so $a^2 \equiv -3 \pmod{83}$, but $\left(\frac{-3}{83}\right) = \left(\frac{-1}{83}\right)\left(\frac{83}{3}\right)(-1)^{(3-1)(83-1)/4} = (-1)\cdot\left(\frac{2}{3}\right)\cdot(-1) = -1$, contradiction. $4^\circ$ We are left with $6n+11 = 53a^2$, $6n+14 = 83b^2$, $20n+19 = 53\cdot83c^2$. Now by some calculations we can get $b \equiv 4 \pmod {53}$ and $a \equiv 5 \pmod {83}.$ Putting $a = 5$ and $b = 4$ we get that $n = 219$ and $c = 1$ with $20n+19 = 4399 = 58\cdot 83$. We know that $a$, $b$ are the smallest then, so it will be the smallest $n = 219$ such that the product is a square. A much faster way to reach a contradiction in the second case is to note that $6n+14=b^2$ implies $b^2\equiv 2 \pmod{3}$, which is impossible. Similarly, in the third case $6n+11=a^2$ implies $a^2\equiv 2 \pmod{3}$ and we have disposed of this case too.

Thx,Alex!..