Color the unit squares black and white in such a way that unit squares with a common side are of different color. Each line goes from a black square to a white square and vice versa; thus whenever the endpoints of a line are in squares of equal color, the line visits an odd number of squares, and otherwise, the line visits an even number of squares. Let $k$ squares out of the ones marked with ring be black. Among the lines drawn by Juku, let $a$ lines have both endpoints in black squares, $b$ lines have both endpoints in white squares, and $c$ lines have endpoints in squares of different color. Then $2a + c = k$ and $a + b + c = n$, implying $a + b = n - k + 2a$. Hence the numbers $a+b$ and $n-k$ have equal parity.
Let the numbers of squares visited by the lines sum up to $s$. Then $s$ can be expressed as the sum of $a+b$ odd numbers and $c$ even numbers, whence $s$ and $a+b$ have equal parity. Consequently, $s$ and $n-k$ have equal parity. Since $n-k$ does not depend on the way Juku draws the lines, $s$ must be always even or always odd.

Every line can be considered as a sequence of unit movements, each having one of four possible directions (right, left, up, down). Thus the number of unit squares visited by a line is $k+1$ where $k$ is the number of unit movements. Let $a, b, c, d$ be the numbers of unit movements of the line in different directions (right, left, up and down, respectively); then the end of the line is located $a-b$ units to the right and $c-d$ units upwards from the beginning of the line. Since $a-b$ and $a+b$ have equal parity, as do $c-d$ and $c+d$, the numbers $k = (a+b) + (c+d)$ and $(a-b) + (c-d)$ have equal parity.
Let the line start in column $x_A$ and row $y_A$ and end in column $x_B$ and row $y_B$. Then $x_B - x_A = a - b$ and $y_B - y_A = c - d$, whence the number of unit squares visited by the line has the same parity as the number $x_B - x_A + y_B - y_A + 1$. But the latter has the same parity as $x_B + x_A + y_B + y_A + 1$. Summing up these numbers for all lines, we obtain that the total number of unit squares visited by the lines has the same parity as the number $s_x + s_y + n$ where $s_x$ and $s_y$ are the sum of all column numbers and all row numbers, respectively, of the squares containing a ring. But this sum does not depend on the way how Juku connects the rings, which proves the desired claim.