To solve the equation

\[
y + (y + 2) + (y + 4) + \ldots + (y + 1996) + (y + 1998) = 1000000,
\]

we first identify the series on the left side.

This series is an arithmetic series where:
- The first term \( a = y \)
- The last term \( l = y + 1998 \)
- The common difference \( d = 2 \)

Now, we need to find the number of terms in this series. The general term of the series can be represented as:

\[
\text{nth term} = y + 2(n-1).
\]

Setting the nth term equal to the last term:

\[
y + 2(n-1) = y + 1998.
\]

Subtract \( y \) from both sides:

\[
2(n-1) = 1998.
\]

Dividing by 2:

\[
n - 1 = 999 \implies n = 1000.
\]

The total number of terms \( n = 1000 \).

Next, we find the sum of the arithmetic series, which is given by the formula:

\[
S_n = \frac{n}{2} (a + l).
\]

Substituting our values:

\[
S_{1000} = \frac{1000}{2} (y + (y + 1998)) = 500 \times (2y + 1998) = 1000y + 999000.
\]

Setting the sum equal to 1,000,000:

\[
1000y + 999000 = 1000000.
\]

Now, isolate \( y \):

\[
1000y = 1000000 - 999000,
\]
\[
1000y = 1000,
\]
\[
y = 1.
\]

Thus, the solution gives:

\[
\boxed{1}.
\]