Suppose this polygon exists, with vertices $A_{1}, A_{2}, \cdots, A_{1990}$. Choose a Cartesian coordinate system, placing vertex $A_{1990}$ at the origin and edge $A_{1990} A_{1}$ along the $x$-axis. From condition (i), we know that each interior angle of this convex polygon is $1-\theta$, where $\theta=2 \pi / 1990$ (why). Considering the edges $A_{1} A_{2}, A_{2} A_{3}, \cdots, A_{1989} A_{1990}, A_{1990} A_{1}$ as vectors, we have (why)
$$\overrightarrow{A_{1} A_{2}}+\overrightarrow{A_{2} A_{3}}+\cdots+\overrightarrow{A_{j} A_{j+1}}+\cdots+\overrightarrow{A_{1989} A_{1990}}+\overrightarrow{A_{1990} A_{1}}=\overrightarrow{0} .$$
Expressing these vectors in complex form, we get (why)
$$\overrightarrow{A_{j} A_{j+1}}=r_{j}^{2} \exp (\mathrm{i} j \theta), \quad 1 \leqslant j \leqslant 1990,$$

where $\mathrm{i}=\sqrt{-1}$, $r_{j}^{2}$ is the length of edge $A_{j} A_{j+1}$, and $A_{1991}=A_{1}$. From equations (1) and (2), we obtain
$$T=\sum_{1 \leqslant j \leqslant 1990} r_{j}^{2} \exp (\mathrm{i} j \theta)=0$$

From condition (2), we know that $r_{1}, r_{2}, \cdots, r_{1990}$ should be a permutation of $1,2, \cdots, 1990$. Thus, our problem is whether and how to select a permutation of $1,2, \cdots, 1990$ such that equation (3) holds.

The sum in equation (3) is typically called an exponential sum. Therefore, we need knowledge about when an exponential sum equals zero. The only known result in this regard is: for any integer $M > 1$,
$$\sum_{1 \leqslant k \leqslant M} \exp (2 \pi \mathrm{i} k / M)=0$$

From this, we can deduce that for any integer $c$ coprime with $M$, we have (why)
$$\sum_{1 \leqslant k \leqslant M} \exp (2 \pi i c k / M)=0$$

To use these conclusions to solve our problem, we need to select $r_{1}, r_{2}, \cdots, r_{1990}$ in a special way so that the sum in equation (3) can be expressed as a combination of the sums in equations (4) and (5). This requires considering how to appropriately combine the terms in the sum in equation (3) corresponding to the angles $j \theta$ to form the sums in equations (4) and (5).

Since the order of the angles $j \theta$ is fixed, this complicates the combination of terms. Given that for any integer $d$,
$$\exp (\mathrm{i} j \theta)=\exp (\mathrm{i}(j+1990 d) \theta)$$

we can find a complete residue system modulo 1990, $u_{1}, u_{2}, \cdots, u_{1990}$, such that (why)
$$T=\sum_{1 \leq j \leqslant 1990} r_{j}^{2} \exp \left(\mathrm{i} \theta u_{j}\right)=0$$

Thus, the problem becomes how to choose $r_{1}, r_{2}, \cdots, r_{1990}$ and $u_{1}, u_{2}, \cdots, u_{1990}$ so that by appropriately combining and summing them, we can achieve our goal. This involves using the properties of residue systems (see Chapter 3, §2). Guided by our approach, noting the property $\mathrm{e}^{\mathrm{i} x}=-1$ and the characteristics of the number 1990, we select $\left\{r_{j}\right\}$ in the form $a_{k, m}=10 k+m(1 \leqslant m \leqslant 10,0 \leqslant k \leqslant 198)$ and $\left\{u_{j}\right\}$ in the form $b_{k, n}=10 k+199 n(0 \leqslant n \leqslant 9,0 \leqslant k \leqslant 198)$. In summing $T$, they are combined as follows:
$$T=\sum_{0 \leqslant k \leqslant 198} T_{k}$$

where
$$\begin{aligned} T_{k}= & a_{k, 1}^{2} \exp \left(\mathrm{i} \theta b_{k, 0}\right)+a_{k, 6}^{2} \exp \left(\mathrm{i} \theta b_{k, 5}\right)+a_{k, 2}^{2} \exp \left(\mathrm{i} \theta b_{k, 2}\right) \\ & +a_{k, 7}^{2} \exp \left(\mathrm{i} \theta b_{k, 7}\right)+a_{k, 3}^{2} \exp \left(\mathrm{i} \theta b_{k, 4}\right)+a_{k, 8}^{2} \exp \left(\mathrm{i} \theta b_{k, 9}\right) \\ & +a_{k, 4}^{2} \exp \left(\mathrm{i} \theta b_{k, 6}\right)+a_{k, 9}^{2} \exp \left(\mathrm{i} \theta b_{k, 1}\right) \\ & +a_{k, 5}^{2} \exp \left(\mathrm{i} \theta b_{k, 8}\right)+a_{k, 10}^{2} \exp \left(\mathrm{i} \theta b_{k, 3}\right) \end{aligned}$$

It is not difficult to see that
$$\begin{aligned} T_{k}= & \exp (10 k \theta \mathrm{i})\left\{\left(a_{k, 1}^{2}-a_{k, 6}^{2}\right) \exp (0 \cdot 2 \pi \mathrm{i} / 5)+\left(a_{k, 2}^{2}-a_{k, 7}^{2}\right) \exp (1 \cdot 2 \pi \mathrm{i} / 5)\right. \\ & +\left(a_{k, 3}^{2}-a_{k, 8}^{2}\right) \exp (2 \cdot 2 \pi \mathrm{i} / 5)+\left(a_{k, 4}^{2}-a_{k, 9}^{2}\right) \exp (3 \cdot 2 \pi \mathrm{i} / 5) \\ & \left.+\left(a_{k, 5}^{2}-a_{k, 10}^{2}\right) \exp (4 \cdot 2 \pi \mathrm{i} / 5)\right\} \\ = & -5 \exp (10 k \theta \mathrm{i})\{(20 k+7) \exp (0 \cdot 2 \pi \mathrm{i} / 5)+(20 k+9) \exp (1 \cdot 2 \pi \mathrm{i} / 5) \\ & +(20 k+11) \exp (2 \cdot 2 \pi \mathrm{i} / 5)+(20 k+13) \exp (3 \cdot 2 \pi \mathrm{i} / 5) \\ & +(20 k+15) \exp (4 \cdot 2 \pi \mathrm{i} / 5)\} \\ = & -5 \exp (10 k \theta \mathrm{i})\{1 \exp (0 \cdot 2 \pi \mathrm{i} / 5)+3 \exp (1 \cdot 2 \pi \mathrm{i} / 5) \\ & +5 \exp (2 \cdot 2 \pi \mathrm{i} / 5)+7 \exp (3 \cdot 2 \pi \mathrm{i} / 5)+9 \exp (4 \cdot 2 \pi \mathrm{i} / 5)\} \\ = & -5 A \exp (10 k \theta \mathrm{i})=-5 A \exp (2 k \pi \mathrm{i} / 199) \end{aligned}$$

Therefore,
$$T=-5 A \sum_{0<k \leq 198} \exp (2 k \pi \mathrm{i} / 199)=0$$

This specifically constructs the required polygon.