Solution:

Let $k$ be the least integer with $k^2 \ge n.$ The function $f$ climbs one step at a time to the next perfect square, so $f(n) = k + (k^2 - n).$ The function $g$ climbs by $2$s, preserving the parity of its argument, and $j^2 \equiv j \pmod 2,$ so $g(n) = j + (j^2 - n)$ where $j$ is the least integer with $j^2 \ge n$ and $j \equiv n \pmod 2.$ If $k \equiv n \pmod 2$ then $j = k$ and the ratio is $1;$ so we need $k \not\equiv n \pmod 2,$ in which case $j = k + 1$ and $g(n) - f(n) = (k+1)^2 - k^2 + 1 = 2k + 2.$

Then $7f = 4g = 4f + 4(2k + 2)$ gives $f = \frac{8(k+1)}{3},$ so $k \equiv 2 \pmod 3$ and $$n = k^2 + k - \frac{8(k+1)}{3},$$ subject to $(k-1)^2 \lt n \le k^2$ and $n \not\equiv k \pmod 2.$

For $k = 2, 5, 8, 11$ the formula gives $n = -2, 14, 48, 100,$ each failing $n \gt (k-1)^2;$ for $k = 14,$ $n = 170$ is in range but has the same parity as $k.$ For $k = 17,$ $n = 289 + 17 - 48 = 258,$ which satisfies $256 \lt 258 \le 289$ and is even while $k$ is odd. Indeed $f(258) = 17 + 31 = 48$ and $g(258) = 18 + 66 = 84,$ with $\frac{48}{84} = \frac{4}{7},$ so the least $n$ is $258.$